The following question is from "The Theory of Finite Groups" by Hans Kurzweil.
Let $G$ be a finite group with $\alpha \in \text{Aut}(G)$ such that $x^{\alpha} \neq x = x^{{\alpha}^2}$ for all $x \neq e$. Then show the following:
a) For every $x \in G$, there exists $y \in G$ such that $x = y^{-1}y^{\alpha}$.
b) $G$ is abelian of odd order.
It's clear that $G$ is of odd order, as all elements except the identity are paired and therefore the order of $G$ is odd, but I cannot prove the other parts. For the first part, I tried $y = x$, $y = x^{\alpha}x$, and $y = xx^{\alpha}$, but none of them seem to work.