Let $G$ be a group. Now consider $\operatorname{Aut} G$ then $\operatorname{Aut} (\operatorname{Aut} G) = \operatorname{Aut}(2G)$, then $\operatorname{Aut}(\operatorname{Aut}(\operatorname{Aut} G))=\operatorname{Aut}(3G)$ and so on.
Let $n$ be the least positive integer for which $\operatorname{Aut} (nG)$ is isomorphic to $\operatorname{Aut}((n+1)G)$. Find the value of $n$ for the symmetric group of degree $6$ and explain why it is so. Just use Automorphism in Group theory.
Note
- There is a finite $n$ for finite group.
- For $S_6$, $\operatorname{Inn}(G)$ has index $2$ in $\operatorname{Aut}(G)$.
Also what I have done that for all other symmetric groups of degree $m\ne 6$, $n=0$. But how do I proceed for $m=6$ after getting $[\operatorname{Aut} S_6:\operatorname{Inn} S_6]=2$?