Composing outer automorphisms with group representations In J.P. Serre's Linear Representations of Finite Groups, two representations $(\rho, V)$ and $(\rho', V')$ of a finite group $G$ are called isomorphic if there exists a linear transformation $T: V \to V'$ such that $T \cdot \rho = \rho' \cdot T$. In the case where $V = V'$, we would call these representations isomorphic if $\rho' = T \cdot \rho \cdot T^{-1} = c_T\circ \rho$, where $c_T$ is conjugation by $T$. Conceptually this makes sense to me because if we identify these two vector spaces using any choice of bases, then this definition essentially says that a change of basis, i.e. how we write the same transformations shouldn't change the representation.
However, if we consider $k$-vector spaces where $GL_n(k)$ has non-trivial outer automorphisms, i.e. automorphisms which are not conjugation by some element, we could call two representations $\rho, \rho':G \to GL_n(k)$ isomorphic if $\rho = \phi\circ \rho'$ for any $GL_n(k)$ automorphism $\phi$. This seems similar and a good candidate because in this case too we can always recover a representation from any isomorphic copies, if we know what the isomorphism is, and because the actions of the group elements maintain their relations under such an isomorphism. An example of this sort of outer automorphism is complex conjugation in $GL_2 (\mathbb{C})$; this is not inner because for any matrix with a non-real determinant, the determinant changes, whereas inner automorphisms must preserve determinants.
I can see that we don't actually define isomorphisms in this fashion because if we did we would lose the uniqueness of the characters up to isomorphism (at the very least). My questions are the following:


*

*If two representations are related in this fashion (i.e. $\rho = \phi \circ \rho'$), what can we say about them?

*If one is irreducible, is the other irreducible too? If so, how exactly do outer automorphisms act on the set of irreducible representations of a fixed dimension?

*Can we calculate the decomposition of one representation using the decomposition of the other? Is there any nice correspondence?

*What does the action of outer automorphisms do at the level of characters? Can this action be described intrinsically on the vector space of class functions?
I think that there are a lot of other questions one could ask in this set-up; any remarks about this situation would be welcome. If there are better answers for similar cases (wherever the question makes sense), for instance locally compact or reductive groups, they would be welcome too.
 A: One way this can happen is if the ground field an automorphism $\psi:k \to k$, this will induce automorphisms $\psi_n: GL_n(k) \to GL_n(k)$ for any $n$ that you can then twist your representations by.
In particular if $K$ is a field extension of $k$ then we get a whole $Gal(K:k)$ worth of such automorphisms to twist things by. If $K = \mathbb{C}$ and $k = \mathbb{R}$ this is just the identity map and complex conjugation.
It turns out that for any $g \in Gal(K:k)$, we get a $k$-linear auto-equivalence of categories from $Rep(G,K)$ to itself (the inverse is just given by $g^{-1}$). In particular it preserves irreducibility, and allows you to match up the decompositions accordingly.
Moreover this equivalence of categories commutes with the forgetful functor to vector spaces over $K$ (and its $k$-linear auto-equivalence induced by $g$).  In particular this means that if you twist a representation by $g \in Gal(K:k)$ and then take characters it is the same as twisting the character values by $g$.
