If $G$ is a nonabelian finite simple group, $Aut(G)$ certainly contains a subgroup isomorphic to $G$, namely $Inn(G)$. Must this be the only subgroup of $Aut(G)$ isomorphic to $G$?


I can prove this in many cases using the smallness of $Out(G)$, but I'm wondering if this is a general fact.



For nonabelian finite simple $G$, does $Aut(G)$ have a unique subgroup isomorphic to $G$?
[Schreier's conjecture](https://en.wikipedia.org/wiki/Schreier_conjecture) (a theorem provable with the classification of finite simple groups) tells you that $Out(G)$ is always solvable. Therefore $Inn(G)$ is in fact the *unique* subgroup isomorphic to $G$ inside $Aut(G)$.


