Let a group $G$ act on a set $S$.
Define the set orbit and pointwise stabiliser for all $A\subseteq S$,
\begin{align*} G\cdot A&:=\{g\cdot a:g\in G,a\in A\}\\ \textrm{Stabp}_G(A)&:=\{g\in G:(\forall a\in A)(g\cdot a = a)\} \end{align*}
Let $X\subseteq S$. Then we have $$\textrm{Stabp}_G(G\cdot X)\trianglelefteq G$$
I would like a source that refers to this, please. I am interested in this and any related results. I'm quite sure it's true - I wrote a proof, but haven't verified it with someone else. If someone could, that would be great.
If the action is faithful, I also wonder if there is some sort of converse statement that can be made? (Normal subgroup $\implies$ ???)
Proof:
By definition, $\textrm{Stabp}_G(G\cdot X)=\{n\in G:(\forall h\in G,x\in X)(n\cdot(h\cdot x)=h\cdot x)\}$.
Let $g\in G$, $n\in\textrm{Stabp}_G(G\cdot X)$. Let $m=gng^{-1}$.
Let $h\in G$, $x\in X$.
Then \begin{align} n\cdot (h\cdot x) = h\cdot x &\iff& (gng^{-1}ghg^{-1})\cdot x &= (ghg^{-1})\cdot x\\ &\iff& (mghg^{-1})\cdot x &= (ghg^{-1})\cdot x \end{align} Note $gGg^{-1} = G$, so letting $k=ghg^{-1}$, we've shown for all $k\in G$, $x\in X$, $$(mk)\cdot x= k\cdot x$$ Hence $m=gng^{-1}\in\textrm{Stabp}_G(G\cdot X)$.
It turns out this result is clear to see if you view the action as a homomorphism into the symmetric group of $S$.
If the action is restricted to a union of orbits $G\cdot X$, then $\textrm{Stabp}_G(G\cdot X)$ is the kernel of the action. That explains why I could not find a formulation of the result in this specific way.
As for a reverse implication, given an arbritrary set $G$ acts on, it is unlikely for all normal subgroups to be found this way. For example, there may only be one orbit while $G$ could have multiple normal subgroups (eg. let $G$ act on itself). So I will abandon this train of thought.