$HDoes anyone know of a counter example or a proof of the following proposition?  If it doesn't hold in general are there any classes of groups for which it holds?

Let $G$ be a non-abelian finite group and let $H<K<G$ with $H$ a maximal subgroup of $K.$  If $gHg^{-1}<K$ then $g^{-1}Hg<K.$ 

Thank you
 A: Consider $G=S_4$ and put $H=\{(1),(12)\}$ and $K=S_3$. Then $H$ is maximal in $K$ because of their orders being $2,6$ respectively, and $2|x|6$ has no nontrivial solutions.
Now let $g=(1432)$. Then [I'm doing compositions left-to-right] $gHg^{-1}=\{(1),(23)\}<K$, however $g^{-1}Hg=\{(1),(14)\}$ is not a subgroup (or even a subset) of $K.$
Note this could be switched around if one wants to do composition in the opposite right-to-left order.
A: The result appears to be false even for $2$-groups.  Here is a counterexample.
Let $\, G\,$ be a Sylow-$2$ subgroup of $\;C_{S_8}((1,2)(3,4)(5,6)(7,8))\;$ (i.e, the centralizer of 
$(1,2)(3,4)(5,6)(7,8)$ in the symmetric group $\,S_8\,$). 
Let $\;K=\langle (1,2)\rangle \times \langle (3,4)\rangle \times\langle (5,6)\rangle$. 
Let $\;H=\langle (1,2)\rangle \times \langle (3,4)\rangle$ and $g=(1,7,5,3)(2,8,6,4)$.
Then (using left-to-right notation):
$gHg^{-1}=(1,7,5,3)(2,8,6,4)(1,2)(1,3,5,7)(2,4,6,8) =(3,4)$, and
$gHg^{-1}=(1,7,5,3)(2,8,6,4)(3,4)(1,3,5,7)(2,4,6,8) =(5,6)$, 
which establishes that $gHg^{-1}\le K$, however 
$g^{-1}Hg\not\le K$, since $$(1,3,5,7)(2,4,6,8)(1,2)(1,7,5,3)(2,8,6,4) =(7,8)\not\in K.$$ 
A: Here is a general counterexample for $p$-groups, $p$ any prime.
Let $\, G\,$ be a Sylow-$p$ subgroup of the centralizer in $S_{p^2}$ of the indicated fixed-point free element of order $p$:
$$\;C_{S_8}((1,2,\dots,p)(p,p+1,\dots,2p) \cdots (p^2-p+1, p^2-p+2, \dots, p^2)$$ 
Let $$\;K=\langle (1,2,\dots,p)\rangle \times \langle (p+1,p+2,\dots , 2p)\rangle \times \cdots \times \langle (p^2-2p+1, p^2-2p+2, \dots, p^2-p)\rangle.$$ 
Let $H$ be the following maximal subgroup of $K$: 
$$\;H=\langle (1,2,\dots,p)\rangle \times \langle (p+1,p+2,\dots , 2p)\rangle \times \cdots \times \langle (p^2-3p+1, p^2-3p+2, \dots, p^2-2p)\rangle.$$  
Let $g=(1,p+1,\dots,p^2-p+1)(2,p+2,\dots,p^2-p+2)\cdots (p,2p,\dots,p^2)$.
Then (using left-to-right notation) $gHg^{-1}\le K$, however $$g^{-1}(1,2,\dots,p)g=(p^2-p+1, p^2-p+2,\dots,p^2)\not\le K.$$  
A: Let $G$ be a group (finite or infinite) with subgroups $H$ and $K$ such that $H$ is maximal in $G$, and $gHg^{-1}\le K$.  Then $g^{-1}Hg\le K$ if and only if either $H=g^{-1}Hg$ or  $K=g^{-1}Kg$.
$(\Leftarrow)$ Obvious.
$(\Rightarrow)$ Suppose $g^{-1}Hg\le K$ and $gHg^{-1}\ne H$. Then $\langle H, gHg^{-1}\rangle=K=\langle H, g^{-1}Hg\rangle$. But $K=\langle H, gHg^{-1}\rangle$ implies that $g^{-1}Kg=g^{-1}\langle H, gHg^{-1}\rangle g=\langle g^{-1}Hg,H\rangle=K$, i.e., $K=g^{-1}Kg$.
A: Relaxing the finite condition, there exists a group $G$ with chain of subgroups $hKh^{-1}<K<h^{-1}Kh$. Which is a kinda cool counter-example (taking $H=hKh^{-1}$ and $g=h^2$). This follows quickly from the groups given as answers to this question.
For example, consider the subgroup $K=\langle a^2\rangle$ of the group $G=\langle a, t; tat^{-1}=a^2\rangle$. Then $tKt^{-1}=\langle a^4\rangle$ and $t^{-1}Kt=\langle a\rangle$, so we get the chain. In fact, this chain continues forever:
$$t^{i+1}Kt^{-(i+1)}<t^{i}Kt^{-i}<\dots<K<\langle a\rangle$$
As I mentioned above, this can be easily made into a formal counter-example by taking $H=tKt^{-1}$ and $g=t^2$. So in the above example, take $K=\langle a^2\rangle$ and $H=\langle a^4\rangle$, and then $g^{-1}Hg<K$ but $gHg^{-1}>K$.
(Note that we take $g:=t^2$ because this moves us two places along the chain, so "jumps" us to the other side of $K$ when moving up the chain. If we took $g:=t$ then we would have $g^{-1}Hg<K$ and $gHg^{-1}=K$)
