Every subgroup is isomorphic to a normal subgroup Let $G$ be a group such that, for every subgroup $H$ of $G$, there exists a normal subgroup $K$ of $G$, such that $H$ is isomorphic to $K$. Under such conditions, can we determine the structure of $G$ ?
This question comes from the discussion of Dedekind group in group theory, which is one such that all of its subgroups are normal.  We know that a Dedekind group is an Abelian
group or a direct product of the quaternion group $Q_8$ and an Abelian group $A$, where $A$ has no elements with order $4$.
So my problem can be regarded as a promotion of Dedekind group that only requires isomorphism. I don't know how to deal with this case.
 A: Let us call $G$ a generalised Dedekind group if every subgroup of $G$ is isomorphic to a normal subgroup of $G$.
As expected, Dedekind groups are generalised Dedekind groups. In addition, YCor has established that a finite generalised Dedekind group $G$ is nilpotent, since its $p$-Sylow subgroups are necessarily normal.
Note that the fundamental group of the Klein bottle, that is, $\mathbb{Z} \rtimes_{-1} \mathbb{Z} = \left\langle a, b \, \vert \, aba^{-1} = b^{-1} \right\rangle$ is an infinite generalised Dedekind group which is not nilpotent.
Here is a family of finite nilpotent groups which are generalised Dedekind groups but not Dedekind groups.

Claim. Let $p$ be a prime number and let $u$ be an integer such that $u \equiv 1 \text{ mod } p \mathbb{Z}$ and $u^p \equiv 1 \text{ mod } p^2 \mathbb{Z}$. Let $$G(p, u) \Doteq \mathbb{Z}/p^2 \mathbb{Z} \rtimes_u \mathbb{Z}/p \mathbb{Z}$$ where the conjugation by $a \Doteq (0, 1 + p \mathbb{Z})$ induces the multiplication by $u$ on $\mathbb{Z}/p^2 \mathbb{Z}$, i.e., $aba^{-1} = b^u$ where $b \Doteq (1 + p^2 \mathbb{Z}, 0)$. Then $G(p, u)$ is a generalised Dedekind group.
If in addition $u \not\equiv 1 \text{ mod } p^2 \mathbb{Z}$, then $G(p, u)$ is not Dedekind.


Proof.
Let $H$ be a subgroup of $G(p, u)$. It is easy to check that $H$ can be generated by two elements $(x, y) = (b^k, b^la^m)$ with $k \in \{0, 1, p\}$ and $l, m \in \mathbb{Z}$.
Let us show first that $G(p, u)$ is a generalised Dedekind group.
If $k = 0$ or $m \equiv 0 \text{ mod } p\mathbb{Z}$ , then $H$ is a cyclic subgroup of order at most $p^2$ and hence isomorphic to either $\{1\}$, $\langle b \rangle$ or $\langle b^p \rangle$ which are normal in $G(p, u)$. Indeed, we have $(b^la^m)^p = 
b^{l(1 + u^m + \dotsb + u^{m(p-1)})}$ with
$1 + u^m + \dotsb + u^{m(p-1)} \equiv p \bmod{p\mathbb{Z}}$, hence the order of $b^la^m$ is at most $p^2$.
If $k = 1$, then $H = \langle b, a^m \rangle$ which is easily seen to be normal in $G(p, u)$.
Let us assume eventually that $k = p$ and $m \not\equiv 0 \text{ mod } p\mathbb{Z}$. Then we have $H = \langle b^p, b^l a^m \rangle$.
If $(b^l a^m)^p \neq 1$, then $b^l a^m$ generates $H$ so that $H$ is isomorphic to a cyclic normal subgroup of $G(p, u)$. Otherwise $\langle b^p \rangle \cap \langle b^l a^m \rangle = \{1\}$. Since $\langle b^p \rangle$ is central in $G(p, u)$, the subgroup $H$ is isomorphic to the Abelian normal subgroup $\langle b^p, a \rangle \triangleleft G(p, u)$.
Thus we have established that $G(p, u)$ is a generalised Dedekind group.
We complete the proof by observing that $G(p, u)$ is Abelian if and only if $u \equiv 1 \text{ mod } p^2 \mathbb{Z}$.

Taking $p = 2$ and $u = 3$ yields the dihedral group of order $8$ which is the smallest finite generalised Dedekind ring which is not Dedekind.
Taking $p = 3$ and $u = 4$ provides us with an example of a $3$-group which is a generalised Dedekind but not Dedekind.

Addendum. Geoff Robinson and YCor have observed in the comments attached to the question  that, more generally, the groups of order $p^3$, for $p$ a prime number, are generalised Dedekind groups. If $p \neq 2$, there are only $2$ isomorphism classes of non-Abelian groups of order $p^3$: the class of $G(p, u)$ (which does not depend on $u$) and the class of the Heisenberg group modulo $p$, that is, $H(\mathbb{Z}/p\mathbb{Z}) \Doteq (\mathbb{Z}/p\mathbb{Z})^2 \rtimes_A \mathbb{Z}/p\mathbb{Z}$ with $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. The proof becomes obvious once we have observed that the centers of these two groups are cyclic of order $p$ and that groups of order $p^2$ are Abelian. If $p = 2$, the latter two classes collapse in one, the isomorphism class of the dihedral group of order $8$, and a new class must be considered, namely the isomorphism class of the  quaternion group.
