Is a finite group with a certain automorphism must be abelian

Question

Let $G$ be a finite group, and let $f:G \rightarrow G$ be an automorphism, such that $x f(x) f(f(x))=1$ for any $x \in G$. Is $G$ must be abelian?

I believe that there are examples where $G$ is not abelian, but I could not find any.

EDIT: I am interested in the case where $f$ has at most 3 fixed elements (including $1$).

Answer 1

Hint: Try to find an example using the simplest automorphism you can think of! ($x\mapsto x$)

Answer 2

To search for a counter example, suppose we have a group $G$ such that the map $g \mapsto g^n$ is an isomorphism for some $n$. (When does this happen? Think about the kernel.)

If this is $f$, the condition you give is equivalent to the condition that the map $g \mapsto g^{n^2+n+1}$ is the identity. That is to say, that the order of any element of $G$ divides $n^2+n+1$.

So to find a counter example with this method, can you find a non-abelian group $G$ and natural number $n$ satisfying the above conditions?

Alternatively, you can start with the smallest non-abelian group, $D_6 (= \langle a,b:a^2=1=b^3, aba=b^{-1}\rangle)$ and try to find an automorphism that satisfies your conditions.

Answer 3

Consider the Heisenberg group mod 3 with the trivial automorphism. it satisfies $x^3=1$ for all $x$ and is not abelian.