Order-conjugate and Hall not implies order-dominating

Statement
There can exist a finite group $$G$$ with a Hall subgroup $$H$$ such that the order and index of $$H$$ in $$G$$ are relatively prime, and such that $$H$$ is conjugate to any subgroup of $$G$$ of the same order as $$H$$, but such that there exists a subgroup math>K of $$G$$ such that the order of $$K$$ divides the order of $$H$$, but $$K$$ is not contained in $$H$$.

Example of the alternating group of degree five
Suppose $$G$$ is the alternating group on the set $$S = \{ 1,2,3,4,5 \}$$. Suppose $$H$$ is the subgroup of $$G$$ that is the alternating group on $$\{ 1,2,3,4 \}$$ (it is isomorphic to particular example::alternating group:A4). In other words, $$H$$ is the stabilizer of the point $$\{ 5 \}$$. Then, $$H$$ is a Hall subgroup of $$G$$. Further, $$H$$ is order-conjugate in $$G$$: the subgroups of the same order as $$H$$ are precisely the stabilizers of points in $$S$$, and these are conjugate to $$H$$ by suitable $$5$$-cycles.

On the other hand, consider the subgroup $$K$$:

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

$$K$$ is a group of order six, isomorphic to the symmetric group of degree three. However, $$K$$ is not contained in any conjugate of $$H$$, because any conjugate of $$H$$ stabilizes some element, and $$K$$ does not stabilize any element.