Series-equivalent abelian-quotient central subgroups not implies automorphic

In terms of series-equivalent subgroups
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) $$G/H$$ (and hence also $$G/K$$) is an abelian group. In other words, $$H$$ and $$K$$ are both fact about::abelian-quotient subgroups, i.e., they both contain the derived subgroup.
 * 3) $$H$$ and $$K$$ are both fact about::central subgroups, i.e., they are both contained in the center of $$G$$.
 * 4) $$H$$ and $$K$$ are not fact about::automorphic subgroups  in $$G$$, i.e., there is no automorphism of $$G$$ sending $$H$$ to $$K$$.

In terms of the second cohomology group
It is possible to have abelian groups $$A$$ and $$B$$ and elements of the fact about::second cohomology group for trivial group action $$H^2(B,A)$$ that are not in the same orbit under the natural action of $$\operatorname{Aut}(B) \times \operatorname{Aut}(A)$$ but nonetheless give isomorphic big groups.

Equivalence of statements
The statement for the second cohomology group can be interpreted in terms of series-equivalent subgroups by identifying $$A \cong H \cong K$$ and $$B \cong G/H \cong G/K$$.

Example
Let $$G$$ be the semidirect product of Z8 and Z8 of M-type (GAP ID: (64,3)) given explicitly by the presentation:

$$\! G := \langle a,x \mid a^8 = x^8 = e, xax^{-1} = a^5 \rangle$$

The center of $$G$$ is $$\langle a^2, x^2 \rangle$$ and it is isomorphic to the direct product of Z4 and Z4. The derived subgroup of $$G$$ is $$\langle a^4 \rangle$$, so any subgroup between these is an abelian-quotient central subgroup.

Let $$H$$ be the subgroup $$\langle a^2, x^4 \rangle$$ and $$K$$ be the subgroup $$\langle a^4, x^2 \rangle$$. Both are central subgroups of $$G$$ isomorphic to particular example::direct product of Z4 and Z2. In both cases, the quotient group is also isomorphic to particular example::direct product of Z4 and Z2. However, there is no automorphism sending $$H$$ to $$K$$ because $$K$$ contains an element whose square is the unique non-identity commutator of $$G$$, while $$H$$ does not.