Let $G$ be an infinite group, $f$ be automorphism with only one fixed point identity. Is $G=\{xf(x^{-1})|x\in G\}$ always true?
I can prove that it is true when $G$ is finite. Construct $g:G\to G$ by $g(x)=xf(x^{-1})$. Then $g(x)=g(y)\implies y^{-1}x=f(y^{-1}x)\implies y^{-1}x=e\implies x=y$; hence $g$ is injective. If $G$ is finite then $g$ onto also, hence done.
But if $G$ is infinite then I cannot prove $g$ is onto. Is there any counter example?
Thanks for assistance in advance.
The non-trivial automorphism of $\mathbb{Z}$, $f:1\mapsto -1$, gives a counter-example. Indeed, writing everything additively, the map $g: x\mapsto x+f(-x)$ defines the homomorphism $x\mapsto 2x$ (homomorphism as the group is abelian). This is non-surjective, as required.