What is the difference between $X^g$ and $\mathrm{Stab}(x)$ in Burnside's Lemma

Question

For some group $G$ acting on a set $X$, Burnside's Lemma states that for the set of orbit of $X$ denoted $O(X)$, then $\forall g \in G$ let $X^g = \{x\in X:g\cdot x = x\}$. How is this set different from the stabilizer $\mathrm{Stab}(X)$?

Answer 1

One is a subset of $X$ and the other a subset of $G$.

Answer 2

Consider where these sets live. By definition, $X^g\subseteq X$. This is the set of all points fixed by a particular $g\in G$. However, $\operatorname{stab}(x) = \{g\in G: g\cdot x = x\}\subseteq G$ for some $x\in X$. This is the set of all group elements that fix a particular point $x$.