I have encounter a question related to automorphism and I found it is really interesting. Unfortunately I cannot completely solve this question.
Question: Suppose that $G$ is a simple nonabelian group. Prove that if $f$ is an automorphism of $G$ such that $x f(x) = f(x) x$ for every $x \in G$ ( i.e. $x$ commute with $f(x)$ ), then $f = 1$ ( the identity automorphism).
Outline of my idea: Let $S = \{a \in G, a^{-1}f(a)\}$ be the group generated by those elements of the form $g^{-1}f(g)$ where $g \in G$, I intend to show that $S$ is a normal subgroup.
For any $b \in G$, let $b = f(c)$ for some $c \in G$. Now: $$ b^{-1}a^{-1}f(a)b = b^{-1}a^{-1}f(ac) = (ab)^{-1}f(ac) = (acc^{-1}f(c))^{-1}f(ac) = $$ $$ [(c^{-1}f(c))^{-1}][(ac)^{-1}f(ac)] \in S $$ which shows that $S$ is a normal subgroup. As $G$ is simple so we must have $S = 1$ or $S = G$. But I have no idea how to rule out the case for $S = G$.
Anyone can help me or give a more simple way to solve this question. Thanks!
This is a well-known problem by Herstein from $1984$. A solution by Thomas J. Laffey is given here:
A Problem by I. N. Herstein (in Monthly)