We consider a Hausdorff topological group $G$ acting on a topological space $X$ [action simply means a continuous map $G\times X\rightarrow X$ verifying $(gh)(x)=g(h(x))$, and $1(x)=x$].
The set $H=\{g\in G : \forall x\in X, g(x)=x\}$ is a normal subgroup of G. For example, it is closed if singletons are closed in $X$ .
I would like to find a Hausdorff topological group $\mathbf{G}$ and a topological space $\mathbf{X}$ such that $\mathbf{H}$ is not closed.
Note: If we do not require $G$ to be Hausdorff there is a plethora of counterexamples.
Let $G$ be a Hausdorff topological group. Let $N$ be a non-closed normal subgroup of $G$. Let $X = G/N$. $G$ acts on $X$. Let $g$ be an element of $G$. We denote by $G_g$ the set $\{h \in G |\ hgN = gN\}$. Then $G_g = gNg^{-1} = N$. Hence $H = \{g\in G : \forall x\in X, g(x)=x\} = \bigcap \{G_g |\ \forall g \in G\} = N$. Hence $H$ is not closed.
For example, let $G = \mathbb{R}$ and $N = \mathbb{Q}$.