TITLE: An automorphism of order 2 which fixes only the identity
QUESTION [3 upvotes]: Let $G$ be a group. Assume there is an element $\phi\in\text{Aut}(G)$ such that $\phi(x)=x\Rightarrow x=e$, where $e$ is the identity of $G$, and that $\phi^2$ is the identity automorphism on $G$. I need to show that $G$ is abelian.
My attempt:
Let $\sim$ be a relation defined on $G$ as follows:
For $x,y\in G$, we write $x\sim y$ iff $x=\phi(y)$ or $x=y$. Then $\sim$ is an equivalence relation on $G$ with each equivalence class containing $2$ elements except for the equivalence class of $e$ which has just $1$ element. So $|G|$ is odd.
Another observation is that if $\phi(x)=y$ then $\phi(xy)=yx$. Since the question asks us to prove that $G$ is abelian, the above suggests that $\phi$ sends each element to its inverse although I am not able to prove this fact. In fact if $\phi$ sends each element to its inverse then it easily follows that $G$ is abelian.
Any help is greatly appreciated. Please try to give insights as to how you arrived at your solution. 
Thank you.

REPLY [4 votes]: Hint: 
Replace $x$ with $x^{-1}y$ in the property of $\phi$. This tells a certain function $G \to G$ is injective.
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