Let $G$ be a group and $M$ be $G$-module. Let $H^1(G,M)$ be first group cohomology.
I heard $G/H$ acts on $H^1(H,M)$ naturally.
What is the standard action ?
I came upon an action * by $(\sigma*f)(g)=\sigma f((\sigma)^{-1}g\sigma)$. Are there other candidates ?
$G/H$ acts naturally on $H^n(H,M)$ for any $n$. Let $M\hookrightarrow I_\ast$ be an injective resolution of $M$ by $G$-modules. $H^n(H,M)$ is the $n$th cohomology of $I_\ast^H$ (valid, since an injective $G$-module is always an injective $H$-module too) and the $G$-action on $I_\ast^H$ obviously is trivial when restricted to $H$, thus inducing an action of the quotient $G/H$ on $I_\ast^H$; each $I_\ast^H$ is naturally a $G/H$ module. Thus $H^n(H,M)$ is naturally a $G/H$-module too. The same argument works for homology.
All of that can be made concrete, however! This is just an explanation for why we should know this to be true with a minimum of thought and without checking anything. Of course, figuring out the precise nature of this action is worthwhile and this is done below. Tl;Dr the action you gave in your post is exactly the right one.
I showcase the analogous action on homology (not cohomology) at the bottom of this post.
With thanks to Mariano Suarez-Alvarez for pointing this out, there should be a very reasonable generalisation of your action to an action on all cocycles. For $g\in G$ and $\phi\in\hom_H(B_n,M)$, $M$ a $G$-module and $B_n$ the normalised bar resolution, $(g\ast\phi)[h_1|\cdots|h_n]:=g\cdot\phi([g^{-1}h_1g|\cdots|g^{-1}h_ng])$ defines a $G$-action which commutes with the boundary maps (this is not too hard to check, the main point is $g^{-1}hh'g=g^{-1}hg\cdot g^{-1}h'g)$ and thus induces a $G$-action on cohomology $H^n(H;M)$.
We would like to check that this in fact is a $G/H$ action. Say $h\in H$. We want, for any $n$-$H$-cocyle $\phi$, $h\ast\phi$ to be cohomologous with $\phi$. In other words, we want the morphism $h\ast(-)-\mathrm{Id}$ to be zero on cohomology. We show this by finding a cochain contraction. Testing low-degree examples I found that:
$$s_n:B_n\to B_{n+1}\\\,[h_1|h_2|\cdots|h_n]\mapsto\\\,[h|h^{-1}h_1h|h^{-1}h_2h|\cdots|h^{-1}h_nh]+\sum_{k=1}^n(-1)^k[h_1|\cdots|h_k|h|h^{-1}h_{k+1}h|\cdots|h^{-1}h_nh]$$
Is a chain contraction of the map of $H$-complexes $h\ast(-)-\mathrm{Id}$, $[h_1|\cdots|h_n]\mapsto h[h^{-1}h_1h|h^{-1}h_2h|\cdots|h^{-1}h_nh]-[h_1|h_2|\cdots|h_n]$.
Thus the dual map $h\ast(-)-\mathrm{Id}$ (which agrees with the previous definition, since if $g\in H$ then $g\cdot\phi([\cdots])=\phi(g\cdot[\cdots]))$ on $\hom_G(B_\ast,M)$ is zero on cohomology, and $h\ast\phi=\phi$ in cohomology for any $h\in H$ and any cocyle $\phi$. Note we need $h\in H$ in order for $h[\cdots]\in B_\ast$ to be well-defined; the general maps $g\ast(-)$ cannot possibly come from a chain map if $g\notin H$ since multiplication by $g$ would be undefined.
We don't actually need an explicit homotopy (see some abstract nonsense below about universal $\delta$-functors) but we can confirm this is a genuine homotopy, using a simplicial framework to simplify calculations:
View $B$ as its underlying simplicial set (with degeneracies $s_i:B_n\to B_{n+1}$ induced by $[h_1|\cdots|h_n]\mapsto[h_1|\cdots|h_i|1|h_{i+1}|\cdots|h_n]$). If the simplicial set-maps $h\ast\phi$ and $\mathrm{Id}$ are homotopic, then they are chain homotopic in homology by the linked post.
I define the simplicial homotopy $H:B\times\Delta^1\to B$ via the following action on the basis simplices (extending in the obvious way to ensure $H$-linearity): $$T_n:B_n\times\Delta^1_n\to B_n\\([h_1|h_2|\cdots|h_n],\tau)\mapsto\begin{cases}[h_1|\cdots|h_{k-1}|h_kh|h^{-1}h_{k+1}h|\cdots|h^{-1}h_nh|]&0<k\le n\\h[h^{-1}h_1h|h^{-1}h_2h|\cdots|h^{-1}h_nh]&k=0\\\,[h_1|h_2|\cdots|h_n]&k=n+1\end{cases}$$Where $k=\min\{0\le j\le n:\tau(j)=1\}$, if the minimum exists, or is $n+1$ in the instance $\tau\equiv0$ identically.
I claim this is genuinely simplicial. It is clearly then a homotopy of the two maps of interest. Moreover, the associated "prism operator" which creates the chain contraction turns out (see the link again) to give exactly the chain homotopy that I declared earlier without proof.
Assuming for the moment that this works, we realise the action on $H^0$ simply takes an invariant $x\in M^H\cong H^0(H;M)$ to $gx$ and it is not too hard to see (following reasoning similar to the below analysis of the $n=1$ case) that this is what happens for my proposed natural $G/H$ action in degree zero. Thus the two automorphisms of $H^\ast(H;M)$ associated to these two different models of $[g]\ast(-)$, for $[g]\in G/H$, are equal in degree zero - and obviously would define natural transformations of the $\delta$-functors since both actions come from (co)chain maps which only use the module structure of $M$, which is preserved by morphisms in our category - and thus are equal in every degree by universality of derived $\delta$-functors. In fact, we can check the action of $H$ is trivial since by the same universality principle; since $h\ast(-)$ is clearly the identity on $H^0(H;M)$, it must be the case that $h\ast(-)$ is the identity in all degrees. The explicit chain homotopy is one way of seeing this (assuming it works for all degrees, which it should) but we know this will be true for abstract reasons.
Therefore, the explicit conjugation action on the bar resolution (which is harder to verify a genuine action) agrees with the natural action coming from more abstract observations. To see how to explore the action as it comes from the more abstract observations, see the next section (which is certainly less general and slick but I guess is a nice manual computation).
In the case $n=1$, this canonical abstract action agrees with the one you propose. Here is an explicit check:
By the acyclic assembly lemma or a spectral sequence argument, there are obvious maps, which happily are quasiisomorphisms, between the $H$-module cocomplexes: $$\hom_H(B_\ast,M)\rightarrow\hom_H(B_\ast,I^\ast)\leftarrow\hom_H(\Bbb Z,I^\ast)\cong(I^\ast)^H$$Where the middle term is the totalisation of an appropriately signed double hom cochain complex. $B$ is the unnormalised bar resolution. These create isomorphisms on cohomology which are the canonical isomorphisms (it turns out they are unique in a sense) relating these two different models of group cohomology, so these are the correct things to consider. This is the principle of $\mathsf{Ext}$ being a balanced functor; you can derive it in either variable and get a canonically isomorphic answer.
Let $\iota$ denote the map $M\hookrightarrow I^0$ and $\partial^1$ the map $I^0\to I^1$. Very importantly, these maps are $G$-module morphisms since we chose $I$ to be a resolution of $G$-modules.
An $H$-derivation $\phi$ on $M$ (identifiable canonically with a cocycle of $\hom_H(B_\ast,M))$ is mapped to a cocycle of $\hom_H(B_\ast,I^\ast)$, namely $(\iota\phi,0)$. By quasiisomorphism, there exists a cocycle $x\in(I^1)^H$ such that $(\iota\phi,0)\simeq(0,\epsilon()\cdot x)$ in cohomology. The canonical $G/H$ action takes $[g]\cdot x:=gx$ (as $x$ is invariant under $H$ this is independent of the representative $g)$ so we need to show $g\ast\phi$ is identified with $gx$ through the above quasiisomorphisms.
Let $z\in I^0\cong\hom_H(B_0,I^0)$ be an element witnessing the equivalence in cohomology, $\partial z=(0,\epsilon()\cdot x)-(\iota\phi,0)=(-\iota\phi,\epsilon()\cdot x)$. That means $\iota\phi(h)=(h-1)z$ and $\partial^1z=x$.
Consider $\iota(g\ast\phi)(h)=\iota(g\phi(g^{-1}hg))=g\iota\phi(g^{-1}hg)=g(g^{-1}hg-1)z=(hg-g)z=(h-1)\cdot gz$. Also, $\partial^1(gz)=g\partial^1z=gx$.
Thus, $gz$ witnesses $g\ast\phi\sim gx$ and your $G/H$-action on $H^1$ is exactly the canonical one.
Now for the action on group homology. We can use the bar resolution to cook up something explicit again.
Let $M$ be a right $G$-module. $H_n(H;M)$ may be computed either by freely resolving $M$ and taking coinvariants or by freely resolving $\Bbb Z$ and tensoring; the two yield the same homology groups by the principle of balanced functors. Let $B_\ast$ denote again the (unnormalised) $H$-bar resolution of $\Bbb Z$.
$H_\ast(H;M)$ is then the homology of the complex $M\otimes_{\Bbb ZH}B_\ast$. It's easier to see how a $G$-action could arise by using some tensor properties; as $B_\ast$ is the free $\Bbb ZH$-module on the set of $\ast$-tuples of elements of $H$ (classically denoted as formal symbols $[h_1|h_2|\cdots|h_n]$ as I've been doing throughout this post) we have: $$M\otimes_{\Bbb ZH}B_n\cong M\otimes_{\Bbb ZH}\bigoplus_{(h_1,\cdots,h_n)\in H^n}\Bbb ZH\cong\bigoplus_{(h_1,\cdots,h_n)\in H^n}M$$As an isomorphism of Abelian groups. The differentials transfer from the left hand side to the right in the obvious way.
Given $g\in G$, we can define a group homomorphism $g\ast(-)$ on $\bigoplus M$ in the following way; for an element $m\in M$ living in the $(h_1,\cdots,h_n)$th summand, map $m$ to $m\cdot g$ in the $(g^{-1}h_1g,\cdots,g^{-1}h_ng)$th summand. Now extend linearly. This clearly defines a right $G$-action. Moreover, you can check (and you should!) that this action commutes with the differentials of the complex. Therefore it induces a right $G$-action on $H_\ast(H;M)$. We would like this to be a right $G/H$-action and again must check $H$ acts trivially i.e. check $h\ast z$ is homologous with $z$ when $z$ is a cycle. But here's the great thing: $m\cdot h$ in the $(h^{-1}h_1h,\cdots,h^{-1}h_nh)$th summand corresponds to $mh\otimes[h^{-1}h_1h|\cdots]=m\otimes h[h^{-1}h_1h|\cdots]$ in the tensor product and we see $h\ast(-)$ is just the result of applying the additive functor $M\otimes_{\Bbb ZH}(-)$ to the map $h\ast(-)$ on $B_\ast$; as we already saw, this map is homotopic to the identity and we get what we want. You could also run the same universal $\delta$-functor argument again if you so wish, being sure you know how to check naturality of the action. And again for the same universality reasons, we know this explicit action is "equal to" the canonical action which comes from abstract observations.
For a refresher, those "abstract observations" are that $H_\ast(H;M)$ is the homology of the complex of coinvariants $F_H$ where $F$ is a free $G$-resolution of $M$ (which will then also be a free $H$-resolution) and $G$ acts naturally on the complex $F_H$, but obviously by construction of the coinvariant groups the action will be trivial when restricted to $H$, so this really is a $G/H$ action (and it would commute with the differentials, be a chain map) so we get a $G/H$ action on homology (and if you used a different free resolution you would get a homotopy equivalence between the two resolutions which would be one of $G$-modules and the $G$-actions would carry over and end up being the same in homology so this really is canonical).
