Order-conjugate and Hall not implies order-dominating

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., order-conjugate Hall subgroup) need not satisfy the second subgroup property (i.e., order-dominating Hall subgroup)
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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</math> of $G$ such that the order of $K$ divides the order of $H$, but $K$ is not contained in $H$.

Proof

Example of the alternating group of degree five

Further information: alternating group:A5, subgroup structure of alternating group:A5

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 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.