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



to test equivalence of representations under automorphism
I assume that $\phi$ is an automorphism of $G.$ Note that if $\phi$ is inner then trivially $\rho$ and $\rho\circ\phi$ are equivalent, thus the answer depends only on the image of $\phi$ in the outer automorphism group $Out(G).$ 


If $G$ is a finite group (or, more generally, compact group) and representations are finite-dimensional, so that they are determined up to isomorphism by their characters, then this problem admits a complete theoretical solution using the character theory. The automorphism group $Aut(G)$ acts on the set $\{C\_i\}$ of the conjugacy classes of $G$, this action factors through the action of $Out(G),$ and


$$\chi\_{\rho\circ\phi}(C)=\chi\_{\rho}(\phi(C)),\qquad (\*)$$


where $\chi\_\rho$ is the character of $\rho$ and $C$ is any conjugacy class. Since representations are determined by their characters, 


$$\rho\circ\phi\simeq\sigma \iff \chi\_{\rho\circ\phi}=\chi\_\sigma,$$


which can be tested using $(\*).$ Of course, applying this method in practice requires good knowledge of the character table of $G$ and the outer automorphism group $Out(G).$


