= Solution
First show $C_N(h)=\{1\}$. If $z\in N$ commutes with $h$ and $z\ne1$, 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 $N$ makes
$$
f_h:N\longrightarrow N,\qquad f_h(y)=h^{-1}y^{-1}hy
$$
well-defined. If $f_h(a)=f_h(b)$, cancellation gives $a^{-1}ha=b^{-1}hb$, and hence $ba^{-1}$ commutes with $h$. Since $ba^{-1}\in N$ and $C_N(h)=\{1\}$, we obtain $a=b$. The map is injective, and an injection of a finite set into itself is surjective. Consequently
$$
\boxed{\text{For every }x\in N\text{ there exists a unique }y\in N\text{ with }[h,y]=x.}
$$
This is the <commutator bijection from a fixed-point-free automorphism>, applied to $\alpha(y)=h^{-1}yh$.
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