Let me start with some definitions and remarks.
I write $.$ to denote a group action of the group $G$ on the set $X$
$O(x) = \{g.x|g \in G\}$ (known as the orbit of $x$)
$X_G = \{x \in X|\forall g \in G: g.x = x\}$
I want to prove that $$\bigcup_{x \in X, |O(x)|=1} O(x) = X_G$$ (if someone knows how to format this better, feel free to comment or edit)
I don't even know whether the statement itself is true (I came up with it myself) but here is my attempt. Can someone verify whether this is correct?
Proof
$\subset$
Let $y \in \bigcup_{x \in X, |O(x)|=1} O(x)$. Then, there is an $x \in X$ such that $y \in O(x)$ and $|O(x)| = 1$. Since $x \in O(x)$, it follows that $x = y$. Hence, $O(x) = \{x\} $ and therefore, we know, by definition of $O(x)$ that for all $g \in G$ we have that $g.x = x$. Hence, $x \in X_G$ and thus $y \in X_G$
$\supset$
Let $y \in X_G$. Then, for all $g \in G$ we have g.y = y. Therefore, $O(y) = \{y\}$. And thus $|O(y)| = 1$. Hence, $y \in \bigcup_{x \in X, |O(x)|=1} O(x) \quad \triangle$