Frobenius kernel implies fixed-point-free automorphism of prime order

Statement
Let $$G$$ be a finite group and $$N$$ be a Frobenius kernel in $$G$$: in particular, $$N$$ is a proper nontrivial complemented normal subgroup with the property that no nontrivial element of $$N$$ commutes with a nontrivial element outside $$N$$. Then, there exists a prime $$p$$ and a fixed point-free automorphism $$\sigma$$ of $$N$$ of order $$p$$.

Proof
Given: A finite group $$G$$, a Frobenius kernel $$N$$ in $$G$$.

To prove: There exists a prime $$p$$ and a fixed point-free automorphism $$\sigma$$ of $$N$$, where $$\sigma$$ has order $$p$$.

Proof: Let $$H$$ be a complement in $$G$$ to $$N$$ (so, $$H$$ is a Frobenius complement). Clearly $$H$$ contains elements of prime order, since it is nontrivial. Pick an element $$g \in H$$ of prime order. Let $$\sigma$$ be the automorphism of $$N$$ induced via conjugation by $$g$$.

If $$a \in N$$ satisfies $$\sigma(a) = a$$, then $$g$$ and $$a$$ commute. By our assumption, no nontrivial element of $$N$$ commutes with any nontrivial element outside $$N$$, so this forces that $$\sigma$$ has no non-identity fixed points.