Series-equivalent abelian-quotient abelian not implies automorphic

Statement
It is possible to have a finite group $$G$$ (in fact, even a group of prime power order) and normal subgroups $$H$$ and $$K$$ of $$G$$ such that:


 * 1) $$H$$ and $$K$$ are fact about::series-equivalent subgroups of $$G$$: $$H$$ and $$K$$ are isomorphic groups and the quotient groups $$G/H$$ and $$G/K$$ are also isomorphic groups.
 * 2) $$H$$ and $$G/H$$ (and hence also $$K$$ and $$G/K$$) are both abelian groups. In other words, $$H$$ is an fact about::abelian normal subgroup as well as an fact about::abelian-quotient subgroup of $$G$$ (and so is $$K$$).
 * 3) $$H$$ and $$K$$ are not fact about::automorphic subgroups in $$G$$, i.e., there is no automorphism of $$G$$ sending $$H$$ to $$K$$.

Weaker facts

 * Series-equivalent not implies automorphic

Example of the nontrivial semidirect product of cyclic groups of order four
We define $$G$$ as the nontrivial semidirect product of Z4 and Z4, explicitly by the presentation:

$$\! G := \langle x,y\mid x^4 = y^4 = e, yxy^{-1} = x^3 \rangle$$

Define $$H$$ as the subgroup $$\langle x^2, y\rangle$$ and $$K$$ as the subgroup $$\langle x,y^2 \rangle$$.

Both $$H$$ and $$K$$ are isomorphic to particular example::direct product of Z4 and Z2, and they are both normal with the quotient group isomorphic to cyclic group:Z2 (because they both have index two).

To see that there is no automorphism of $$G$$ sending $$H$$ to $$K$$, note that the only non-identity squares of elements in the two subgroups are $$y^2$$ (in $$H$$) and $$x^2$$ (in $$K$$) respectively. An automorphism of $$G$$ sending $$H$$ to $$K$$ must therefore send $$y^2$$ to $$x^2$$. This is not possible since $$x^2$$ is in the derived subgroup of $$G$$ (it is the only non-identity commutator) and $$y^2$$ is not.