 Commutator bijection from a fixed-point-free automorphism

Let $N$ be a finite group and $\alpha$ an automorphism whose only fixed element is the identity. If $\alpha (a{)}^{-1}a=\alpha (b{)}^{-1}b$, then $\alpha (b{a}^{-1})=b{a}^{-1}$, so $a=b$. The displayed map is therefore a bijection. For conjugation by a nonidentity element of a Frobenius complement, it gives the commutator map $y\mapsto h^{-1}{y}^{-1}hy$.