Order-conjugate not implies order-dominated

Statement
It is possible to have a finite subgroup $$H$$ of a group $$G$$ such that $$H$$ is conjugate to any subgroup of $$G$$ of the same order as $$H$$, but such that there exists a subgroup $$K$$ of $$G$$ whose order is a multiple of the order of $$H$$, but such that $$H$$ is not conjugate to any subgroup of $$K$$.

Related facts

 * Order-conjugate not implies order-dominating
 * Order-conjugate and Hall not implies order-dominating

Example of the alternating group of degree five
Suppose $$G$$ is the alternating group on the set $$\{ 1,2,3,4,5 \}$$. Suppose $$H$$ is the subgroup of $$G$$ given by:

$$H := \{, (1,2,3), (1,3,2), (1,2)(4,5), (2,3)(4,5), (1,3)(4,5) \}$$.

$$H$$ is conjugate to any other subgroup of $$G$$ of the same order.

On the other hand, suppose $$K$$ is the stabilizer of $$\{ 5 \}$$ in $$G$$. Then, $$K$$ is the alternating group on $$\{ 1,2,3,4 \}$$ and is isomorphic to the alternating group of degree four. The order of $$K$$ is $$12$$, which is a multiple of the order of $$H$$. However, no conjugate of $$H$$ is contained in $$K$$, since $$H$$ does not stabilize any element, but every conjugate of $$K$$ stabilizes some element.