The existence of a group automorphism with some properties implies commutativity.

Let $G $ be a finite group, $T$ be an automorphisom of $ G $ st $ Tx = x \iff x=e $. Suppose further that $ T^2 =I $. Prove that $ G $ is abelian.

I was thinking if I show $ T aba^{-1} b^ {-1}=aba^ {-1}b^{-1} \forall a, b \in G$. But I was unable to show it. Please give me any hints about it.

HINT: Consider $\sigma:x\mapsto x^{-1}T(x)$.

Note that, in general, if $f(x)$ is a fixed point free automorphism of $G$ (fixed point free means $x=1\Leftrightarrow f(x)=x$), then $\sigma(x)=x^{-1}f(x)$ is a bijection from $G$ to $G$ (though this function is not in general a homomorphism).

Hint: $|G|<\infty$, $T$ is an automorphism with the given property then $\forall g\in G$ can be written is of the form $g=x^{-1}T(x)$ to prove this result just define $f(x)=x^{-1}T(x)$ and show that $f$ is onto.

Now come to your probelm:

By above result $\forall a\in G$ we can write $a=x^{-1}T(x)$ for some $x\in G$ so $T(a)=T(x^{-1})T^2(x)=T(x^{-1}) x=[x^{-1}T(x)]^{-1}=a^{-1}$ so $T(ab)=(ab)^{-1}=b^{-1}a^{-1}$ so $T(a)T(b)=b^{-1}a^{-1}$ so $a^{-1}b^{-1}=b^{-1}a^{-1}$ so $ab=ba$, so $G$ is abelian.