show that orbits have the same order under the normal subgroup of a transitive group 
Let $G$ be a group that works transitively on $X$, and let $N$ be a normal subgroup of $G$. Show that the orbits of $X$ under $N$ have the same order, that is $\operatorname{ord}(Nx)=\operatorname{ord}(Ny)$ for all $x,y\in X$.

I'm not sure how to show this. I was thinking of either showing a bijection, maybe
$$
\phi\colon Nx\to Xy\colon g\circ x\mapsto g\circ y,
$$ or somehow using the orbit-stabilizer theorem. Either way I'm stuck. The reason I can't make anything out of $\phi$ is because I don't even know how to show injectivity to start with, though I was hoping that the fact that $N$ is normal would somehow aid in that;
$$
g\circ x=h\circ x\implies h^{-1}g\circ x=x.
$$
I wonder if I could maybe use the following: choose for $x\neq y\in X$, an element $g\in G$ such that $g\circ x=y$. We know then that
$$
G_y=G_{g\circ x}=g G_x g^{-1}.
$$
Any hints?
 A: Yes, just what you stated.
The automorphism of $G$ $z\mapsto g z g^{-1}$ maps $N$ to $N$ ( since $N$ is normal)  and $G_x$ to $G_{gx}=G_{y}$, so induces a bijection from $N/(N\cap G_x)$ to $N/(N\cap G_y)$.
A: By transitivity, there exists $g\in G$ such that $gx=y$. Define the map $\phi: Nx\to Ny$ by $\phi(n x)=gng^{-1}y$ for all $n\in N$. Note that $gng^{-1}\in N$ by normality of $N$.
To prove injectivity, assume $gn_1g^{-1} y=\phi(n_1x)=\phi(n_2x)=g n_2 g^{-1}y$. Cancelling the $g$ both sides, we get $n_1g^{-1}y=n_2g^{-1}y$. Since we have a group action, $n_1x=n_1g^{-1}y=n_2g^{-1}y=n_2x$.
To prove surjectivity, if $ny\in Ny$, then take $\phi(g^{-1}ngx)=g(g^{-1}ng)g^{-1}y=ny$. Note that $g^{-1}ng\in N$ by normality.
A: You're really close to the solution. As you mentionned, there is $g \in G$ such that $g \circ x=y$. Now
\begin{align}
N(y)=N(g \circ x) &= \{hg \circ x \ : \ h \in N \} \\
&= \{\ g(g^{-1}hg)\circ x \ : \ h \in N \} \\
&= \{gk \circ x \ : \ k \in N \} \\
&= gN(x)
\end{align}
Now $|N(x)| = |gN(x)|$ (There is an obvious bijection between the 2).
