TITLE: Normalizer of group action
QUESTION [8 upvotes]: Take a group $G$ acting faithfully on a set $X$, and let $H \leq G$. It can be easily shown that the elements of $N_{G}(H)$ stabilize the collection of orbits of $H$ (as a set, ie orbits are mapped to orbits).
Is the converse true? That is, if we take $\mathcal{O}$ to be the collection of orbits of $H$, do we always have that $\mathrm{Stab}_{G}(\mathcal{O}) = N_{G}(H)$? (I'm happy if everything is assumed to be finite, but more general answers are also welcome.)
Edit: Since, as mentioned in the answer of runway44, this can be considered by looking at what happens with each of the $G$-orbits on $X$, I would like to know if this is true for $G$ acting transitively and faithfully on $X$.

REPLY [2 votes]: I wanted to extend the analysis of runway44's answer to the case where $G$ acts transitively on $X$, and give an example that shows that even here, the answer is negative.
If we assume that $G$ acts faithfully and transitively on $X$, then we can put $K = \mathrm{Stab}_{G}(x)$ for some arbitrary $x \in X$; then there is a natural correspondence between $X$ and $G/K$, the cosets of $K$ (in contrast to runway44's answer, I am going to be working with right cosets, and a right group action). So each element of $X$ corresponds to a coset $Kg$ for some $g \in G$, and an $H$-orbit on $X$ corresponds to the double-coset $KgH$.
Now, if some $n \in G$ stabilizes the $H$-orbits on $X$, then for each $g \in G$ we have
$KgHn = Kg^{\prime}H$
for some $g^{\prime} \in G$. Since $gn \in KgHn$, we have
$KgHn = KgnH$. On the other hand, we can write
$KgHn = Kgn(n^{-1}Hn)$ giving us $KgnH = Kgn(n^{-1}Hn)$; since $gn$ covers all the elements of $G$ as $g$ does, this means
$$KgH = Kg(n^{-1}Hn)$$
for all $g \in G$, or equivalently, 
$$x^{H} = x^{n^{-1}Hn}$$ for all $x \in X$. In other words, the most we can conclude is that $H$ and $n^{-1}Hn$ have the same orbits on $X$.
From this, we can see that even when $G$ acts faithfully and transitively on $X$, it is possible that $N_{G}(H) < \mathrm{Stab}_{G}(O(H))$. For example, let $G = S_{6}$ acting naturally on $\{1,2,3,4,5,6\}$, and take $H = \langle (1,2,3)(4,5,6),\ (1,2) \rangle$. If $n = (1,4)(2,5)(3,6)$, then $n$ stabilizes the orbits of $H$. But 
$$n^{-1}Hn = \langle (1,2,3)(4,5,6),\ (4,5)\rangle \ne H.$$