Intermediately characteristic not implies isomorph-containing in group of prime power order

Statement
For any prime number $$p$$, there exists a finite $$p$$-group $$G$$ and a subgroup $$H$$ of $$G$$ such that $$H$$ is an intermediately characteristic subgroup of $$G$$ but is not an isomorph-containing subgroup (and hence not an fact about::isomorph-free subgroup, or fact about::isomorph-free subgroup of group of prime power order) of $$G$$.

In fact, we can obtain an example where $$H$$ is a characteristic maximal subgroup of $$G$$. Thus, a characteristic maximal subgroup of a group of prime power order need not be an fact about::isomorph-free maximal subgroup of group of prime power order.

Proof
For any $$p$$, set $$G$$ to be SmallGroup(p^4,4) and $$H$$ to be the unique characteristic subgroup of order $$p^3$$. $$H$$ is a direct product of a cyclic group of order $$p^2$$ and a cyclic group of order $$p$$. All the other maximal subgroups of $$G$$ are also isomorphic to $$H$$.

For instance, for $$p = 2$$, set $$G$$ to be SmallGroup(16,4) and $$H$$ to be the unique characteristic subgroup of order eight. Explicitly:

$$G := \langle a,b,c \mid a^2 = b^4 = e, b^2 = c^2, ab = ba, ac = ca, cbc^{-1} = ab \rangle, \qquad H = \langle b^2, bc \rangle$$.

Both the other subgroups of order eight are isomorphic to $$H$$, and all are isomorphic to the direct product of Z4 and Z2.