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 $G$?
Best Answer
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..