Characteristic not implies sub-(isomorph-normal characteristic) in finite

Statement
We can have a finite group $$G$$ and a characteristic subgroup $$H$$ of $$G$$ such that $$H$$ is not sub-(isomorph-normal characteristic) in $$G$$. In other words, there is no ascending chain of subgroups from $$H$$ to $$G$$ such that each member is an isomorph-normal characteristic subgroup of its successor.

Related facts

 * Characteristic not implies sub-isomorph-free in finite
 * Characteristic not implies isomorph-free in finite

The center of a non-abelian group of odd prime cube order
Let $$p$$ be an odd prime. Let $$G$$ be the non-abelian group of order $$p^3$$ and exponent $$p$$. Let $$H$$ be the center of $$G$$. Then, we have:


 * $$H$$ is characteristic in $$G$$.
 * $$H$$ is not sub-(isomorph-normal characteristic) in $$G$$: In fact, no proper subgroup of $$G$$ containing $$H$$ is isomorph-normal and characteristic. $$H$$ itself is not isomorph-normal, because there are other cyclic groups of order $$p$$. Moreover, $$H$$ is a maximal characteristic subgroup of $$G$$, so there is no other possibility.