An automorphism that has no fixed points except for the identity and is its own inverse implies commutativity. Why do $x$ and $f(x)$ commute? 
Let $G$ be a finite group and suppose there exists $f\in\text{Aut}(G)$ such that $f^2=\text{id}_G$, i.e., $f$ is its own inverse, and such that $f$ has no fixed points other than the identity $e$ of $G$, i.e., $f(x)=x\Rightarrow x=e$. Show that $G$ is necessarily abelian.

This question has been asked here. I could understand the somewhat standard solution of using the bijection $x\mapsto xf(x)^{-1}$, but I was searching for a more natural way to solve this prblem. The OP of that question said, "Also, it's easy to see that $g$ and $f(g)$ commute." This is exactly what I wanted to prove in my attempt, but I could not prove it. How does it follow?
 A: You are right that it is not easy to see that $x$ and $f(x)$ always commute. Indeed what is easy to see is that $x$ and $f(x)$ cannot commute unless $f(x)=x^{-1}$: applying $f$ to the product $xf(x)$ gives $f(x)x$, so saying that $x$ and $f(x)$ commute means that $xf(x)$ is a fixed point of $f$, which by hypothesis means it equals $e$, so $f(x)=x^{-1}$. So showing that $x$ and $f(x)$ commute means proving that $f$ is the map $x\mapsto x^{-1}$ which is well known to be a morphism only if $G$ is Abelian, so showing this amounts to solving the linked question (which is what the accepted answer there simply remarks, but it fails to question the reasoning that led to commutation of $x$ and $f(x)$).
So bravo for asking this question. Ultimately the answer to your title question is that $x$ and $f(x)$ commute because it can be shown that $f(x)=x^{-1}$ for all$~x$ (as show in the other answer to linked question), and these two group elements always commute.
So the user that posed the linked question either made a mistake in saying "it is easy to see that $g$ and $f(g)$ commute" (which I take to mean, substantially easier than to solve the given problem outright), or maybe in a flash of insight (s)he saw a simple argument for this commutation that escapes us. It is always difficult to show the latter cannot be the case, but here at least we can be sure there is no proof using only group elements and the axioms of group theory, because mentioned commutation fails without the hypothesis that $G$ is finite. (Explicitly using finiteness of $G$ requires a switch from elements to sets and cardinalities, as you can see in the proof by Bruno Joyal, which uses that an injective map from a finite set to itself is always surjective; this fails for infinite sets.) And we know the commutation fails in the infinite case because of the example given in that same answer: take $G=\langle x,y\rangle$ a free group on $2$ generators (whose elements can be given as words in $x,x^{-1},y,y^{-1}$, and $f$ the automorphism induced by interchanging $x$ and $y$ (which acts by systematically replacing $x\leftrightarrow y$ and $x^{-1}\leftrightarrow y^{-1}$), for which $x$ and $f(x)=y$ do not commute. (It is easy to see that $f^2=\mathrm{id}_G$; that $f(z)=z$ implies $z=e$ is a bit harder, but follows from the that that elements of a free group are uniquely represented by reduced words, which are words that never have a letter and its inverse right next to each other.)
