Subgroup property stronger than being characteristic In what follows, all groups are assumed to be finite.  
Recall that if $K \leq H \leq G$ are groups, $K$ is said to be a weakly closed subgroup of $H$ in $G$ if, for all $g \in G$, $g^{-1}Kg \leq H$ implies that $g^{-1}Kg = K$.
The subgroup property I am interested in is this one: what kind of subgroup of $H$ is $K$ when, for all groups $G$ with $H \leq G$, $K$ is a weakly closed subgroup of $H$ in $G$?
It is trivial to verify that $1$ and $H$ are subgroups of $H$ with this property. (Here, already, the finiteness condition is important: if $H$ is infinite, one could have $g^{-1}Hg < H$ instead of having $g^{-1}Hg = H$. So I won't go there.)  
It is clearly necessary that $K$ be a characteristic subgroup of $H$ for this unnamed property to hold:
If $H$ is finite, so is its holomorph $Hol(H) = H \rtimes Aut(H)$. So let $G = Hol(H)$. Then, regarding $Aut(H)$ as a subgroup of $G$, let $\alpha \in Aut(H)$. Then $\alpha^{-1}K\alpha \leq H$ because $\alpha$ is an automorphism of $H$. Since $K$ is, by assumption, a weakly closed subgroup of $H$ in $G$, this means $\alpha^{-1}K\alpha = K$. Since $\alpha \in Aut(H)$ was arbitrary, it follows that $K$ is a characteristic subgroup of $H$.  
On the other hand, merely being characteristic is not sufficient:
Let $G = <(1243675), (4657)(23)>$, so that $G \cong GL_{3}(2)$. Let $H = <(4657)(23), (67)(23)>$ and $K = <(45)(67)>$. Then $K = Z(H)$, so $K$ is a characteristic subgroup of $H$. It is also true that $H \leq G$, and that $g = (13)(57) \in G$.
But then $g^{-1}Kg = <(47)(65)> \leq H$, but $<(47)(65)> \neq < (45)(67) >$.
I have heard of fully invariant subgroups, but this also shows that a fully invariant subgroup need not have this property: in the example above, $Z(H)$ is a fully invariant subgroup of $H$ because $Z(H)$ is the subgroup generated by the squares in $H$.  
What I can say is that if a subgroup $K$ of $H$ is the only subgroup of $H$ with elements having the orders they do, then $K$ is this kind of subgroup of $H$. (So, in the above example, $<(4657)(23)>$ is the unique cyclic subgroup of order 4 in $H$ and it is this kind of subgroup of $H$, unlike $Z(H)$.)
Does this kind of subgroup have a name? What other sets of conditions on a characteristic subgroup imply this stronger property?
 A: There must be no other subgroup of $H$ isomorphic to $K$. Certainly this property implies your property, as $g^{-1}Kg$ is always isomorphic to $K$. Conversely if $K'\leq H$ is isomorphic to  $K$ let $\alpha:K\to K'$ be an isomorphism, and let $G$ be the HNN extension $H\ast_\alpha$. The natural map $H\to G$ is an injection, and $K$ and $K'$ are conjugate in $G$.
If moreover $H$ is finite then $G$ is residually finite. Thus there is a finite quotient $G'$ of $G$ such that the composition $H\to G\to G'$ is injective. Since $K$ and $K'$ were conjugate in $G$ they remain so in $G'$.
A: Instead of going via HNN extensions, one can work entirely with finite groups
to show that if $X \cong Y$ are subgroups of a finite group $H$, then there exists a finite group $G \supseteq H$ such that $X$ and $Y$ are conjugate in G.
The trick is to take $G$ to be the full symmetric group on he elements of $H$,  with $H$ embedded as the permutations induced by right multiplication (as in Cayley's theorem). Let $\theta$ be an isomorphism from $X$ to $Y$. Now choose a set $T$ of representatives for the left cosets of $X$ in $H$, and similarly, a set $S$ of representatives for the left cosets of $Y$. Note that
$|T| = |H:X| = |H:Y| = |S|$, and let $\sigma$ be an arbitrary bijection from $T$ to $S$. Extend $\sigma$ to a permutation of $H$ by setting
$(tx)^\sigma = t^\sigma x^\theta$.
Now $\sigma$ is an element of $G$, and I argue that conjugation in $G$ by $\sigma$ carries right multiplication by $x$ to right multiplication by $x^\theta$. This will show that the copies of $X$ and $Y$ in $G$ are conjugate in $G$.
Write $y = x^\theta$ and let $r_x$ and $r_y$ be the corresponding right multiplication maps on $H$. I want to show that $\sigma^{-1}r_x\sigma = r_y$, or equivalently, that $r_x\sigma = \sigma r_y$.  To check this, we apply each side to an arbitrary element $h$ in $H$. Applying $r_x\sigma$ yields $(hx)^\sigma$, and applying $\sigma r_y$ yields $h^\sigma y$. We thus want $(hx)^\sigma = h^\sigma y$. Now write $h = tu$, with $t \in T$ and $u \in X$. Then
$$\
(hx)^\sigma = (tux)^\sigma = t^\sigma(ux)^\theta = t^\sigma u^\theta x^\theta = (tu)^\sigma y = h^\sigma y ,
$$
as wanted.  
