Isomorph-automorphic subgroup

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Definition

Symbol-free definition

A subgroup of a group is termed isomorph-automorphic if given any other isomorphic subgroup, there is an automorphism of the whole group, mapping the subgroup isomorphically to the other one.

Definition with symbols

A subgroup $H$ of a group $G$ is termed isomorph-automorphic if whenever there exists a subgroup $K$ of $G$ such that $H$ and $K$ are isomorphic groups, there exists an automorphism $\sigma$ of $G$ such that $\sigma(H) = K$ .

Examples

In a finite cyclic group, every subgroup is isomorph-automorphic (in fact, every subgroup is isomorph-free: no two subgroups are isomorphic).

Similarly, in a finite elementary Abelian group, every subgroup is isomorph-automorphic. That's because given any two subspaces of a finite-dimensional vector space that have the same dimension, there is an automorphism of the whole space taking one to the other.

By Sylow's theorem, every Sylow subgroup is isomorph-automorphic.

On the other hand, many subgroups are not isomorph-automorphic:

In $\mathbb{Z}$ , the group of integers, any nontrivial subgroup is of the form $m\mathbb{Z}$ , $m \ne 0$ , hence is isomorphic to $\mathbb{Z}$ . However, there is clearly no automorphism of $\mathbb{Z}$ mapping it to a proper subgroup.
Any infinite-dimensional vector space is isomorphic to a subspace of codimension one, but there is no automorphism mapping the whole space to such a subspace.
In the direct product $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$ , the direct factor $\mathbb{Z}/2\mathbb{Z}$ is isomorphic to the subgroup of multiples of 2 in the direct factor $\mathbb{Z}/4\mathbb{Z}$ . However, there is no automorphism taking the first to the second.
In the dihedral group of order eight, the center is of order two, and any subgroup generated by a reflection has order two, so they are isomorphic. However, there is no automorphism taking the center to a subgroup generated by a reflection.