Here is my proof of the first part.
(=>). Let $G$ act on a set $X$ and for some $x\in X$. In order to show that $G_x$ is a normal subgroup of $G$, we will show that for all $g \in G$ we have $gG_x = G_xg$. Let $g$ be any element of $G$ and $y = gx$, an element in the orbit of $G(x)$. We know from a previous result that points in the same orbit have conjugate stablizers, i.e. $gG_xg^{-1}=G_y$, and since $G_x=G_y$, we have $gG_xg^{-1}=G_x$ or $gG_x = G_xg$. The stabilizer is normal.
(<=). Suppose $G_x \lhd G$ for some $x \in X$ and let $y \in G(x)$. I have no idea where to go from here.
Best Answer
($<=$)
Since $y\in G(x)$ we have $y=g.x$ for some $g\in G$.
Then $$h\in G_y\iff g^{-1}hg\in G_x\tag1$$ or equivalently:$$G_y=gG_xg^{-1}\tag2$$
Then the normality of $G_x$ tells us that $G_x=G_y$.