Retract not implies normal complements are isomorphic

Statement
Suppose $$G$$ is a group and $$H$$ is a fact about::retract of $$G$$. In other words, $$H$$ has a fact about::normal complement in $$G$$: there exists a normal subgroup $$N$$ of $$G$$ such that $$NH = G$$ and $$N \cap H$$ is trivial.

Then, there may exist another normal complement $$M$$ to $$H$$ in $$G$$ such that $$M$$ is not isomorphic to $$N$$.

Example of the dihedral group
Consider the following example:


 * $$G$$ is the dihedral group of order eight:

$$\langle a,x \mid a^4 = x^2 = e, xax^{-1} = a^{-1} \rangle$$.


 * $$H$$ is the subgroup $$\{ x, e \}$$, i.e., a two-element cyclic subgroup.
 * There are two normal complements to $$H$$ in $$G$$. The subgroup $$N = \{ x, x^2, x^3, e \}$$ is cyclic of order four and the subgroup $$M = \{ e, x^2, ax, a^3x \}$$ is a Klein-four group.