TITLE: for every $\sigma\in \rm Aut(G)$ and for every $x\in G$, $\sigma(x)=x$ or $\sigma(x)=x^{-1}$
QUESTION [5 upvotes]: Let $G$ be a group such that for every $\sigma\in \rm Aut(G)$ and for every $x\in G$, $\sigma(x)=x$ or $\sigma(x)=x^{-1}$.
I want to prove that $G$ is solvable.

REPLY [2 votes]: In such a group, every automorphism $\sigma$ is trivial or has order $2$. Hence $\operatorname{Aut}(G)$ is abelian. In particular, $\operatorname{Inn}(G) \cong G / Z(G)$ is abelian, and thus $G$ is solvable.