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 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 Abelian normal subgroup (?) as well as an Abelian-quotient subgroup (?) of $G$ (and so is $K$).
3. $H$ and $K$ are not Automorphic subgroups (?) in $G$, i.e., there is no automorphism of $G$ sending $H$ to $K$.

Related facts

Stronger facts

Statement	Constraint on $G,H,K$	Smallest order of $G$ among known examples	Isomorphism class of $G$	Isomorphism class of $H,K$	Isomorphism class of quotient group $G/H, G/K$
series-equivalent not implies automorphic in finite abelian group	$G$ is also a finite abelian group	128	direct product of Z8 and Z4 and V4	direct product of Z8 and V4	direct product of Z4 and Z2
series-equivalent abelian-quotient central subgroups may be distinct	$H$ and $K$ are both central subgroups of $G$ (in other words, the quotient actions are both trivial.	64	semidirect product of Z8 and Z8 of M-type	direct product of Z4 and Z2	direct product of Z4 and Z2

Weaker facts

• Series-equivalent not implies automorphic

Proof

Example of the nontrivial semidirect product of cyclic groups of order four

Further information: nontrivial semidirect product of Z4 and Z4, subgroup structure of nontrivial semidirect product of Z4 and Z4

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 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.