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Let $G$ be a finite group and let $\phi$ be an automorphism of $G$. We define an action of $\operatorname{Aut}(G)$ on the set $\operatorname{Rep}(G)$ of complex-valued representations of $G$ by ${}^\phi\rho(g) := \rho(\phi(g))$. Let $[\operatorname{Rep}(G)]$ denote the set of isomorphism classes of representations of $G$, where $\rho \cong \rho'$ iff there exists an invertible linear map $T$ such that $T^{-1}\rho(g)T = \rho'(g)$ for all $g \in G$.

If $\phi$ is an inner automorphism of $G$, then the induced action of $\phi$ on $[\operatorname{Rep}(G)]$ is trivial. Thus we get an action of the outer automorphism group $\operatorname{Out}(G)$ on $[\operatorname{Rep}(G)]$.

My question: is this action faithful, i.e. can there be outer automorphisms of $G$ that fix all isomorphism classes of representations of $G$?

Shaun
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gimothytowers
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    To expand on how Jeremy Rickard's link answers the question, a representation is determined up to isomorphism by its character so if an automorphism maps all elements to conjugates it fixes all representations (since it fixes their characters). (Conversely, since the characters separate conjugacy classes of $G$, an automorphism that fixes all representations must send each element of $G$ to a conjugate.) – Eric Wofsey May 21 '24 at 14:42
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    I would say that Eric's comment shows that while the linked question answers this completely, it is actually not a duplicate and requires a bit more thought. – Steve D May 21 '24 at 16:06
  • I might be misunderstanding the linked question, but I think I require something stronger. It's not enough that for all $g \in G$ there exist a $T_g$ such that $\rho(\phi(g)) = T_g\rho(g)T_g^{-1}$ -- I would like this $T$ to be the same for all $g$. – gimothytowers May 21 '24 at 16:31
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    @gimothytowers See Eric Wofsey's comment. A complex representation of a finite group is determined up to isomorphism by its character, and so if $\rho'(g)=T_g\rho(g)T_g^{-1}$ for each $g$, then $\rho'$ and $\rho$ have the same character and hence are isomorphic (and so there is a single $T$ that will do). But I take the point that there is this extra step to reduce this to the linked question, and I have voted to reopen so that somebody can post the argument as an answer. – Jeremy Rickard May 21 '24 at 19:11

1 Answers1

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Suppose an automorphism $\phi$ has the property that for each $g\in G$, $\phi(g)$ is a conjugate of $g$. Then for any representation $\rho$, the character $\chi_\rho$ satisfies $\chi_\rho(\phi(g))=\chi_\rho(g)$, since characters are constant on conjugacy classes. Thus $\chi_\rho=\chi_{{}^\phi\rho}$ and hence $\rho\cong {}^\phi\rho$ since representations with the same character are isomorphic.

Conversely, if $\phi(g)$ is not a conjugate of $g$ for some $g\in G$, then there is a representation $\rho$ such that $\chi_\rho(\phi(g))\neq\chi_\rho(g)$ (since characters of representations span the set of class functions on $G$, so they separate distinct conjugacy classes). In particular, this implies $\rho\not\cong {}^\phi\rho$.

So, an automorphism of $G$ fixes all isomorphism classes of representations of $G$ iff it sends each element of $G$ to a conjugate. There exist finite groups which have outer automorphisms which send each element to a conjugate; see Example of an outer automorphism that maps all elements to conjugates..

Eric Wofsey
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