Let $G$ act transitively on a set X and let $x\in X$. Prove that the fixed set $X^{G_x}$ is a block of $X$. Deduce that if G acts primitively, then $X^{G_x}={x}$ or else $G_x=\{1\}$ and $X$ is a finite set with $|X|$ prime.
Here, $G_x$ is a stabilizer of element $x\in X$.
I can't solve this problem. $X^{G_x}\cap gX^{G_x}$ for some $g\in G$ makes me frustrated. So I can't show that $X^{G_x}\cap gX^{G_x}=\emptyset$ or $gX^{G_x}= X^{G_x}$ for all $g\in G$.
I even can't solve the deduce part. I understand that if G acts primitively, then $X^{G_x}={x}$ or else $G_x=\{1\}$, but I can't prove that $|X|$ is prime.
Help me to solve this problem!
Here are some hints. For the main problem, show that the relation $x \sim y \Leftrightarrow G_x=G_y$ is an equivalence relation on $X$. Then $X^{G_\alpha}$ is a $\sim$-equivalence class and these classes are permuted by the action of $G$.
For the second question, if $G_x=\{1\}$, then, for every subgroup $H\le G$ and $x \in X$, the orbit $Hx$ of $x$ under $H$ is a block. So $G$ is primitive iff it has no proper nontrivial subgroups i.e. iff $|G|=|X|$ is prime.