Isomorph-freeness is not upper join-closed

Statement
We can have a situation where a subgroup $$H \le G$$ is isomorph-free in two intermediate subgroups $$K$$ and $$L$$, but is not isomorph-free in $$\langle K, L \rangle$$.

Example of a non-Abelian prime-cubed order group
Let $$p$$ be an odd prime, and let $$G$$ be a non-Abelian group of order $$p^3$$ obtained as the semidirect product of a cyclic group $$A$$ of order $$p^2$$ and a cyclic group $$B$$ of order $$p$$. Let $$H$$ be the center of $$G$$. Let $$K$$ and $$L$$ be two cyclic subgroups of order $$p^2$$. Then:


 * $$H$$ is isomorph-free in $$K$$ as well as in $$L$$: it is the cyclic subgroup of order $$p$$ in the cyclic group of order $$p$$.
 * $$H$$ is not isomorph-free in the join $$\langle K, L \rangle = G$$: The cyclic subgroup $$B$$ is isomorphic to $$H$$, for instance.