Let $G$ be a group and $S$ a $G$-set with action $(g,s) \mapsto gs$. For some $s \in S$, let the stabilizer of $s$, $G_s=\{g \in G\,|\,gs=s\}$ be normal in $G$. What does this let us say about the action of $G$ on $S$?
I thought it might be interesting to look at an action of $G/G_s$ on $S$. However, something like $(gG_s,s) \mapsto gs$ isn't even well-defined in general.
Are there situations in which we can recover anything interesting?
On
First, assume the group action is transitive. Then all stabilizers are conjugate to one another. Therefore, if the stabilizer is normal, then all elements of $S$ have the same stabilizer. This is equivalent to saying that if $g$ fixes one point of $s$, then it fixes them all. This means that the induced action of $G/G_s$ on $S$ is regular: For every $s,t \in S$, there is a unique $g \in G/G_s$ such that $gs=t$. In particular, $|G/G_s| = |S|$.
Now even if the action is not transitive, then you can still say all of the above about the orbit of $s$, a point for which $G_s$ is normal.
What normality of the stabiliser says is exactly that every group element $g\in G$ that fixes $s$ also fixes the entire orbit $Gs$ pointwise. Conversely any $g\in G$ that fixes any element of the orbit $Gs$ will also fix $s$.
These two parts are equivalent, although the first sentence says that every conjugate of $G_s$ contains $G_s$, while the second sentence says that any conjugate of $G_s$ is contained in $G_s$. Even though a subgroup $H$ may strictly contain a conjugate ${}^gH$ of itself, if it contains all its conjugates then it must be equal to them all, in other words normal (should $H\supsetneq {}^gH$ then ${}^{g^{-1}}H\supsetneq H$, and the hypothesis excludes this).