Isomorph-freeness is not finite-intersection-closed

Verbal statement
An intersection of isomorph-free subgroups of a group need not be isomorph-free.

Statement with symbols
It is possible to have subgroups $$H,K \le G$$ such that both $$H$$ and $$K$$ are isomorph-free (in other words, there is no other subgroup of $$G$$ isomorphic to $$H$$ and no other subgroup isomorphic to $$K$$) but $$H \cap K$$ is not isomorph-free.

Example of a dihedral group
Let $$G$$ be the dihedral group of order $$24$$: in other words, $$G$$ is the semidirect product of the cyclic group $$C$$ of order $$12$$ and the cyclic group of order two, acting via the inverse map. Let $$H$$ be the subgroup generated by multiples of $$2$$ in $$C$$, and $$K$$ be the subgroup generated by multiples of $$3$$.

Both $$H$$ and $$K$$ are isomorph-free subgroups of $$G$$: $$H$$ is isomorphic to the cyclic group of order six and $$K$$ is isomorphic to the cyclic group of order four. To see that they are isomorph-free, observe that within the cyclic group of order twelve, there are no elements of order six (respectively four) other than those in $$H$$ (respectively $$K$$) while all elements outside $$C$$ have order two.

On the other hand, the intersection $$H \cap K$$, which comprises the multiples of six, is a cyclic group of order two, and is isomorphic to the cyclic groups of order two generated by any element outside $$C$$. Hence, $$H \cap K$$ is not isomorph-free.