 Solution

First show $C_N(h)=\{1\}$. If $z\in N$ commutes with $h$ and $z\ne 1$, then $z\notin H$ because $N\cap H=\{1\}$. But $z^{-1}hz=h$, so the nonidentity element $h$ lies in $H\cap H^z$, contradicting the Frobenius complement property. Thus conjugation by $h$ has no nonidentity fixed element in the Frobenius kernel.

Normality of $N$ makeswell-defined. If $f_h(a)=f_h(b)$, cancellation gives $a^{-1}ha=b^{-1}hb$, and hence $b{a}^{-1}$ commutes with $h$. Since $b{a}^{-1}\in N$ and $C_N(h)=\{1\}$, we obtain $a=b$. The map is injective, and an injection of a finite set into itself is surjective. ConsequentlyThis is the commutator bijection from a fixed-point-free automorphism, applied to $\alpha (y)=h^{-1}yh$.