Prove that $\ker(\rho) = \bigcap_{x \in X} stab_G(x)$ 
Let $\sigma_g: X \to X,\hspace{1mm} x \mapsto g \ast x$. The resulting function $\rho: G \to \text{Sym}(X),\hspace{1mm} g \mapsto \sigma_g$ is called the permutation representation of the action of $G$ on $X$.
Prove that $\ker(\rho) = \bigcap_{x \in X} stab_G(x) = \{g \in G: g \ast x = x \text{ for every } x \in X \}$

Assume that $a \in \bigcap_{x \in X} stab_G(x)$ and let $x \in X$. Then $a \ast x = x$ for all $x \in X$. The corresponding permutation is $\sigma_a(x) = a \ast x = x$, which is the identity permutation in $\text{Sym}(X).$ Therefore $\rho(a) = \sigma_a = 1$, which means that $a \in \ker(\rho)$. Thus $ \bigcap_{x \in X} stab_G(x) \subseteq \ker(\rho)$.
Now let $a \in \ker(\rho)$. Then $\rho(a) = 1$ and $\sigma_a = 1: X \to X, x \mapsto a \ast x = x$ for every $x \in X$ (Equivalently, $\sigma_a(x) = x$ for all $x \in X$). This implies that $a \in \bigcap_{x \in X} stab_G(x).$ Thus $\ker(\rho) \subseteq \bigcap_{x \in X} stab_G(x).$ Combined with the fact that $\bigcap_{x \in X} stab_G(x) \subseteq \ker(\rho)$, it follows that $\ker(\rho) = \bigcap_{x \in X} stab_G(x)$

Method. You need to prove that $\ker \rho \subset \bigcap \mathrm{Stab}_G(x)$. Take $a \in \ker \rho$, then pick $x \in X$ and check if $a \in \mathrm{Stab}_G (x)$. But going back to the definition of $\rho$, his should be nearly obvious...
Solution. Let $a,x$ as I just stated. $a \in \ker \rho$, so $\rho(a)$ is the null element of $\mathrm{Sym}(X)$, which is the identity function on $X$. Therefore for every $x \in X$, we have $\rho(a)(x) = x$, which means (remember the definition of $\rho$ !) taht $a*x = x$. Therefore $a \in \mathrm{Stab}_G(x)$.
