Krdytkyu

Let $f$ be an automorphism of the finite group $G$ such that $fcirc f=id$ and $f(x)=ximplies x=e$

Prove that $f(x)=x^{-1}~forall xin G$

If we can prove that $f(x)$ and $x$ commute for any given $xin G$ then we're done with the proof because it would imply

$$xf(x)=f(x)x$$

$$f(f(x))f(x)=f(x)x$$

Because $fcirc f=id$. And since $f$ is a homomorphism:
$$f(f(x)x)=f(x)ximplies f(x)x=eimplies f(x)=x^{-1}$$

I don't know how to proceed

The fact that $G$ is abelian can be deduced once we have $f(x)=x^{-1}$ because that function is a homomorphism if and only if $G$ is abelian

Exercise: Let $G$ be a finite group , $f$ an automorphism of $G$ with $f(x) =x iff x=e$. Prove that every $g in G$ can be written as $$g=x^{-1}f(x)$$

Proof : Define $alpha:G to G$ by $alpha(x)=x^{-1}f(x)$. Use the hypothesis to prove $alpha$ is one one. That is, $$alpha(x)=alpha(y) Longrightarrow x^{-1}f(x)=y^{-1}f(y)$$ $$hspace{2cm}Longrightarrow yx^{-1}=f(yx^{-1})$$

$$hspace{2.2cm}Longrightarrow yx^{-1}=e;;text{by hyp}$$

$$Longrightarrow y=x$$

Finiteness implies $alpha$ is onto as well. Onto means, for every $g in G$, there exist $x in G$ such that $alpha(x)=x^{-1}f(x)=g$ $;blacksquare$

Proof of your result: $$f(g)=fBig(x^{-1}f(x)Big)=f(x^{-1})f^2(x)=f(x^{-1})x=Big[x^{-1}f(x)Big]^{-1}=g^{-1}$$