Inconsistency between a regular action by automorphisms and non-cyclicality

Question

Let $p$ be a prime. Quite tautologically, $G=\operatorname{Aut}(C_p)$ acts by automorphisms on $\mathbb Z/p\mathbb Z$, with orbits $\{0\}$ and $X:=\{1,\dots,p-1\}$. Let $\varphi$ be the action. This means that, for every $k\in X$, there is unique $g\in G$ such that: $$\varphi_g(1)=k \tag1$$ From $(1)$ follows that: $$k^{|g|}=1 \tag2$$ For $p=5$, if $\operatorname{Aut}(C_5)$ were isomorphic to $C_2\times C_2$, then by $(2)$ we'd have $k^2=1$ for every $k=1,2,3,4$, which is clearly false (in particular for $k=2,3$). Therefore, $\operatorname{Aut}(C_5)\cong C_4$.

Q: Can we generalize this argument to show that, for every prime $p$, $\operatorname{Aut}(C_p)$ cannot contain a subgroup isomorphic to $C_q\times C_q$ (where $q$ is a prime such that $q^2\mid p-1$), and hence it is cyclic?

Answer 1

(Turning the comment into an answer.)

Your proof boils down to showing that if the automorphism group of $\mathrm{Aut}(C_5)$ were not cyclic, then the polynomial $x^2-1$ over $\mathbb{F}_5$ would have more than $2$ roots, which is impossible. This is essentially the same argument that one uses in general to show that if $k$ is a field then any finite subgroup of $k^\times$ is cyclic. The answer I linked to in the comment does this, but actually this answer is even closer to your original argument, since it more explicitly shows that any finite subgroup of $k^\times$ can't have a subgroup isomorphic to $C_q\times C_q$ for any prime $q$, as you discuss in the question. In any case, once you know that any finite subgroup of $k^\times$ is cyclic, then you recover your claim by observing that $\mathbb{F}_p^\times$ can be identified with $\mathrm{Aut}(C_p)$ (by looking at where an automorphism takes $1$).