What are all finite groups such that all isomorphic subgroups are identical?

Let $G$ be a finite group with this property. Let $a\in G$ and let $a$ have order $k$. Then every element of order $k$ in $G$ generates $\left<a\right>$. So there are exactly $\varphi(k)$ elements of order $k$ in $G$. So for each $k$, $G$ has either $0$ or $\varphi(k)$ elements of order $k$.

Let $|G|=n$. Then for each $k$ for which there are elements of order $k$ in $G$, then $k\mid n$. But $n=\sum_{k\mid n}\varphi(k)$. If for some $k\mid n$ there are no elements of order $k$ in $G$, then the total number of elements in $G$ is $<n$, which is impossible.

In particular, $G$ has an element of order $n$. Thus $G$ is cyclic.

From the comments, here is short answer.

As pointed out, any exception to the conjecture must be a group that is non-abelian /and/ has every subgroup normal.

This means the group has the property of being a Dedekind group. In particular, there is a result that the quaternion group $Q_8$ is a subgroup of all such groups. The quaternion group as 6 elements of order four, which cannot belong to a single cyclic group of order four.

We conclude such an exception is not possible.

This depends heavily of the properties of a Dedekind group, which I have no insight to. A better answer would be much appreciated.