TITLE: to test equivalence of representations under automorphism
QUESTION [6 upvotes]: Hi all,
I would like to propose the following problem:
Given two representations $\rho$ and $\tau$ of a group $G$ over complex number, we would like to know if there exists an automorphism $\phi$, such that $\rho\circ\phi$ and $\tau$ are equivalent. 
Is there any mathematical results concerning this problem? It seems that to understand the action of automorphisms on the set of irreducible representations is crucial.
Thank you!
Youming

REPLY [3 votes]: Victor's answer shows that it is important to understand the action of $Out(G)$ on the conjugacy classes of $G$. This can be interesting even in the abelian case, where the problem amounts to calculating the orbits of $Aut(G)$ on $G$. Even though the answer was well-known, we found a combinatorial approach to the question quite interesting.
The idea (which can be formulated for any group) is to define a pre-order on elements: $x\leq y$ if an endomorphism maps $y$ to $x$. The equivalence classes associated to this pre-order ($x\sim y$ if $x\leq y$ and $y\leq x$) are unions of outer equivalence classes. In the finite abelian case, these classes are precisely the $Aut(G)$-orbits.
The question of what other classes of groups this method can be applied to remains open.