# Isomorphic quotients by isomorphic normal subgroups

In this recent question, Iota asked if, given a finite group $G$ and two isomorphic normal subgroups $H$ and $K$, it would follow that $G/H$ and $G/K$ are isomorphic. This is not true (a simple example given by $G=\mathbb{Z}_2\oplus\mathbb{Z}_4$, with $H=\langle (1,0)\rangle$ and $K=\langle (0,2)\rangle$). A sufficient condition to guarantee isomorphic quotients is the existence of an automorphism $\varphi\in\mathrm{Aut}(G)$ such that $\varphi(H)=K$.

One can also have $G/H\cong G/K$, but $H$ and $K$ not isomorphic. For example, take again $G=\mathbb{Z}_2\oplus\mathbb{Z}_4$, and take $H=\mathbb{Z}_2\oplus\langle (0,2)\rangle$, isomorphic to the Klein $4$-group, and $K=\{0\}\oplus\mathbb{Z}_4$. Then $G/H\cong G/K\cong \mathbb{Z}_2$.

And of course, it is trivial to have $H\cong K$ and $G/H\cong G/K$.

Now, here’s the question:

Question. Can we have a finite group $G$, normal subgroups $H$ and $K$ that are isomorphic as groups, $G/H$ isomorphic to $G/K$, but no $\varphi\in\mathrm{Aut}(G)$ such that $\varphi(H) = K$?

It is not hard to come up with examples with infinite $G$. For example, take

let $H=2G$, and let $K$ be the subgroup generated by $H$ and the generator of the cyclic factor $\mathbb{Z}_2$. Then both $H$ and $K$ are isomorphic to a direct sum of countably many copies of $\mathbb{Z}_2$, as are the quotients $G/H$ and $G/K$. But $H$ is a verbal subgroup, hence fully invariant, so any automorphism of $G$ maps $H$ to $H$, so there does not exist any $\varphi\in\mathrm{Aut}(G)$ such that $\varphi(H)=K$.

Does someone have an example with finite $G$? Note that I am not requiring that $\varphi$ be a lift of the given isomorphism of $G/H$ with $G/K$.