Complements to normal subgroup need not be automorphic

Statement
Suppose $$G$$ is a group, $$N$$ is a normal subgroup, and $$H$$ and $$K$$ are permutable complements to $$N$$ in $$G$$. Then, it is not necessary that there exists an automorphism of $$G$$ sending $$H$$ to $$K$$.

Related facts

 * Schur-Zassenhaus theorem
 * Complements to Abelian normal subgroup are automorphic
 * Complement to normal subgroup is isomorphic to quotient

A generic example
Let $$A$$ be any non-Abelian group. Consider $$G = A \times A$$ and the subgroup $$N = A \times \{ e \}$$. Let $$H$$ be the subgroup $$\{ e \} \times A$$ and $$K$$ be the subgroup $$\{ (a,a) \mid a \in A \}$$.

Note that:


 * $$N$$ is normal in $$G$$: In fact, it is a direct factor of $$G$$.
 * $$H$$ is a permutable complement to $$N$$ in $$G$$.
 * $$K$$ is a permutable complement to $$N$$ in $$G$$.
 * $$H$$ is normal in $$G$$: In fact, it is a direct factor of $$G$$.
 * $$K$$ is not normal in $$G$$: Pick $$a,b \in A$$ such that $$a,b$$ do not commute. Then, we have $$(b,e)(a,a)(b,e)^{-1}) = (bab^{-1},a)$$. Thus, a conjugate of an element in $$K$$ lies outside $$K$$.