I have a short question: if $G$ acts on $X$, is the intersection of all the stabilisers the same as the conjugates of the stabilisers? In other words does, for any $z\in X:$
$$\bigcap_{g\in G} g\,\text{stab}(z)\,g^{-1}=\bigcap_{x\in X} \text{stab}(x)?$$
No, this is not the case. As a counterexample, consider any action such that $G$ fixes some point of $X$ but acts non-trivially on another.
If the action of $G$ is transitive, however, then your statement is true, for in this case all point stabilizers are conjugate to each other. Try showing that $$ \operatorname{stab}(g\cdot x) = g \operatorname{stab}(x) g^{-1}. $$