Characteristic-isomorph-free not implies normal-isomorph-free in finite

Statement
We may have a group with a characteristic subgroup such that there are no other characteristic subgroups isomorphic to it, but there are other normal subgroups isomorphic to it.

Example of an Abelian group
Let $$p$$ be any prime, and let $$G := \mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p^2\mathbb{Z}$$. In other words, $$G$$ is a direct product of cyclic groups of order $$p$$ and $$p^2$$ respectively.

Consider the subgroup $$H := \operatorname{Agemo}^1(G)$$: the set of all elements of $$G$$ that can be written as $$p^{th}$$ powers. This set is $$0 \times p\mathbb{Z}/p^2\mathbb{Z}$$.


 * $$H$$ is a characteristic subgroup: It is defined as the set of $$p^{th}$$ powers, so it is clearly invariant under conjugation.
 * There is no other characteristic subgroup of $$G$$ of order $$p$$: All the subgroups of order $$p$$ in $$G$$ lie inside $$\Omega_1(G) = \mathbb{Z}/p\mathbb{Z} \times p\mathbb{Z}/p^2\mathbb{Z}$$. Further, automorphisms of the form $$(a,b) \mapsto (a + b,b)$$ permute all the other subgroups.
 * There are other normal subgroups of $$G$$ of order $$p$$: In fact, there are $$p$$ of them: the other $$p$$ subgroups of $$\Omega_1(G)$$.

Thus, $$H$$ is characteristic-isomorph-free in $$G$$ but is not normal-isomorph-free in $$G$$.