Characteristic not implies characteristic-isomorph-free in finite

Statement
We can find a group $$G$$ (in fact, we can choose $$G$$ to be a finite group) and characteristic subgroups $$H,K$$ of $$G$$ that are not isomorphic to each other but distinct.

Related facts

 * Characteristic-isomorph-free not implies normal-isomorph-free
 * Normal-isomorph-free not implies isomorph-free
 * Characteristic equals characteristic-isomorph-free in finite abelian group

Example of the infinite cyclic group
Let $$\mathbb{Z}$$ be the infinite cyclic group: the group of integers under addition. Then, $$n\mathbb{Z}$$ is a characteristic subgroup of $$\mathbb{Z}$$ for any $$n$$, and all the $$n\mathbb{Z}$$ are isomorphic for $$n$$ a nonnegative integer. Thus, there are distinct characteristic subgroups that are isomorphic.

Example of a finite solvable group
Let $$A$$ be a nontrivial metabelian group that is also a centerless group, with derived subgroup $$B$$. Let $$C$$ be a group isomorphic to $$B$$. Then, in the direct product $$G := A \times C$$, we have:


 * The subgroup $$H = \{ e \} \times C$$ equals the center of $$A \times C$$, hence is characteristic.
 * The subgroup $$K = A \times \{ e \}$$ equals the derived subgroup of $$A \times C$$, hence is characteristic.

Thus, we have two distinct characteristic subgroups that are isomorphic.

A particular case of this is where $$A$$ is the symmetric group on three letters. In this case, $$B$$ is A3 in S3 and $$C$$ is cyclic of order three.

Example of a finite nilpotent group
There are groups of order $$p^5$$ with characteristic maximal subgroups that are isomorphic. For instance, the group with GAP Group ID $$(243,5)$$ has three pairwise isomorphic characteristic subgroups (each with group ID $$(81,3)$$).