Show that a finite group with certain automorphism is abelian 
Let $G$ be a finite group and $f:G\to G$ an isomorphism. If $f$ has no fixed points (i.e., $f(x)=x$ implies $x=e$) and if $f\circ f$ is the identity, then $G$ is abelian. (Hint: Prove that every element in $G$ has the form $x^{-1}\cdot f(x)$.)

With the hint, I can see that for all $t\in G$, $f(t)=t^{-1}$, and since $f$ is an isomorphism, $G$ is abelian. But I can't see how to prove the hint or why it's evident.
Source : Rotman J.J. Introduction to the theory of groups, exercise 1.50
 A: Metahint (hint for the hint): Show that if $x^{-1}f(x) = y^{-1}f(y)$, then $x=y$. Conclude that the map (not necessarily a group homomorphism, at this point it's just a set-theoretic function) $x\mapsto x^{-1}f(x)$ from $G$ to itself is one-to-one. 
Note that the two things you don't use once you have the hint are the property that $f(z)=z$ implies $z=e$, and the condition that $G$ is finite. I wonder if that will play a role in establishing the hint...?

Additional notes. The conditions that $G$ be finite and that $f$ have no fixed points are both necessary for the conclusion to follow.
An example of an infinite nonabelian group $G$, an isomorphism $f\colon G\to G$ with no fixed points, and with $f\circ f = \mathrm{id}_G$ is given by letting $G$ be the free group of rank $2$, freely generated by $x$ and $y$, and $f\colon G\to G$ be the map that swaps $x$ and $y$. It has no fixed points, since a reduced word that begins with $x$ or $x^{-1}$ has image that begins with $y$ or $y^{-1}$, and viceversa; composing the map with itself gives the identity; but $G$ is not abelian. Even though the (set theoretic) map $w\mapsto w^{-1}f(w)$ is still one-to-one, it is no longer onto: the image contains only words where the sum of the exponents is equal to $0$, so for example $x$ cannot be written as $w^{-1}f(w)$ for any $w\in G$. 
For an example with $G$ finite but not abelian if $f$ has fixed points, simply take any nonabelian group with a noncentral element $g$ of order $2$ (any nonabelian simple group will do), and $f$ to be conjugation by $g$. 
A: For proving the second part, that is $f \circ f = I$ implies $G$ is abelian you can proceed as follows:
Consider $g \in G$. By the above $\mathsf{Hint}$ every $g \in G$ can be written as $g = x^{-1} f(x)$ for some $x \in G$. 
We then have 
\begin{align*} g = x^{-1} f(x) \Longrightarrow f(g) &= f(x^{-1}f(x))= f(x^{-1}) \cdot f(f(x)) \\ &=(f(x))^{-1} \cdot x = \bigl[x^{-1} f(x)\bigr]^{-1}=g^{-1}
\end{align*}
Now we prove that if $f: G \to G$ defined by $f(g) = g^{-1}$ is a Automorphism if $G$ is abelian. 
To see this note that $f(ab)=(ab)^{-1} = b^{-1} \cdot a^{-1} $. But $$f(ab) = f(a) \cdot f(b) \Rightarrow f(ab) = a^{-1}b^{-1}$$ From above we have $$b^{-1}a^{-1} = a^{-1}b^{-1} \Rightarrow (ab)^{-1} = (ba)^{-1} \Rightarrow ab=ba$$ Hence $G$ is abelian.
