Let $G$ be a permutation group of a set $X\neq \emptyset$ and let $x,y\in X$. We define:
\begin{align*}&G_x:=\{g\in G\mid g(x)=x\} \\ &G_{x\rightarrow y}:=\{g\in G\mid g(x)=y\} \\ &B:=\{y\in X\mid \exists g\in G: g(x)=y\}\end{align*}
I have shown that $G_x$ is a subgroup of $G$ als that the set $\{G_{x\rightarrow y}\mid y\in B\}\subseteq 2^G$ is a partition of $G$.
Let $g\in G_{x\rightarrow y}$ and let $u\in G_x$ and $v\in G_{x\rightarrow y}$ and so it holds that $g\circ u\in G_{x\rightarrow y}$ and $g^{-1}\circ v\in G_x$. The maps \begin{align*}\alpha:G_x\rightarrow G_{x\rightarrow y}, \ u\mapsto g\circ u \\ \beta: G_{x\rightarrow y}\rightarrow G_x, \ v\mapsto g^{-1}\circ v\end{align*} are to each other inverse bijections.
Let $G$ be finite. I want to show that that $|G|=|B|\cdot |G_x|$. Could you give a hint?
We have that $\{G_{x\rightarrow y}\mid y\in B\}\subseteq 2^G$ is a partition of $G$. Does this mean that $|G|=|\{G_{x\rightarrow y}\mid y\in B\}|$ ?
If it is like that, then we have to show that $|\{G_{x\rightarrow y}\mid y\in B\}|=|B|\cdot |G_x|$, right?
It is also given a hint that we can use the idea that we use when we show that there are $48$ symmetries of a cube.
Consider the map of sets: \begin{gather} \varphi: G/G_x \longrightarrow B, \qquad gG_x\longmapsto g(x) \end{gather} We want to show that $\varphi$ is well defined and that is a bijection.
$\varphi$ is well defined: If $g_1G_x = g_2G_x$, then $g_2^{-1}g_1\in G_x$. Hence $g_2^{-1}g_1(x)=x$ that is $g_1(x)=g_2(x)$. So we have \begin{equation} \varphi(g_1G_x)= g_1(x) = g_2(x) = \varphi(g_2G_x) \end{equation} and $\varphi$ is well defined.
$\varphi$ is injective: If $\varphi(g_1G_x) = \varphi(g_2G_x)$, we have $g_1(x) = g_2(x)$, that is $g_2^{-1}g_1(x)=x$. Hence $g_2^{-1}g_1 \in G_x$ that is $g_1G_x=g_2G_x$ and $\varphi$ is injective.
$\varphi$ is surjective: Take an element $y\in B$, then by definition there exists $g\in G$ such that $g(x)=y$. So: \begin{equation} \varphi(gG_x) = g(x) = y \end{equation} and $\varphi$ is surjective.
Now $G$ is finite so we can pass to the cardinality and we have: $$ \frac{|G|}{|G_x|} = |B| \Leftrightarrow |G| = |G_x|\cdot|B| $$
If you are familiar with group actions, the bijection showed is the connection between the stabilizer of an element $x\in X$ (the subgroup $G_x$) and the its orbit (the set $B$).