# Let group $G$ acts on set $M$. Why are there always $\# G_m$ elements in $G$ map $m$ into any $n$ in the orbit $G\cdot m$?

Suppose that $$G$$ is a finite group, and let $$\# G$$ denote the number of elements in $$G$$. Let $$m$$ be a point of $$M$$, and consider the orbit of $$m$$ under $$G$$, which contains $$\#(G\cdot m)$$ elements. If $$n$$ is an element of this orbit, then $$n=am$$ for some $$a$$ in $$G$$. If $$n=bm$$ as well, then $$a^{-1}b$$ must lie in $$G_m$$. This means that to each element $$n$$ there are exactly $$\#G_m$$ group elements which map $$m$$ into $$n$$. Therefore, since every element of $$G$$ carries $$m$$ into some orbit element, we conclude that

$$\#G=\#(G\cdot m)\#G_m$$

Above is an excerpt from Sternberg, Group Theory and Physics. To be clear, $$G\cdot m \subset M$$ is the orbit of $$m\in M$$, and $$G_m \subset G$$ the isotropic group of each $$m$$. I know that for any two elements $$a,b \in G$$ that map $$m$$ to $$n$$, there are two elements $$a^{-1}b$$ and $$b^{-1}a$$ in $$G_m$$, but I don't understand the emphasized bit, since $$n$$ is any element in $$G\cdot m$$, while $$\#G_m$$ is the fixed number. We can't be sure if there exists $$b$$ in $$G$$ besides $$a$$ such that $$bm=n$$, and even if with this particular $$n$$ there are $$\#G_m$$ ways to map $$m$$ to $$n$$, then nothing precludes another $$n'\in G\cdot m$$ from having a different number of ways to be mapped from $$m$$.

So why is that?

What the text says is that if $b$ maps $m$ to $n=a\cdot m$, then $b\in aG_m$. In fact the converse is true : if $b=ag$ for some $g\in G_m$, then $$b\cdot m=(ag)\cdot m=a\cdot (g\cdot m)=a\cdot m=n.$$ Thus the set of elements of $G$ that maps $m$ to $n$ is exactly $aG_m$, which has the same number of elements as $G_m$. This is true for any $n$ in the orbit of $G$ (although of course with a different element in the orbit you will get a different coset of $G_m$); in particular, the number of ways to map $m$ to any element of its orbit only depends on $m$.
For any $n$ in the orbit of $m$, there obviously exists $a \in G$, and $a \notin G_m$, such that $am = n$. Now, note that $aG_mm = am = n$. Thus, you have exactly #$G_m$ elements that map $m$ to $n$ (note that $G_m$ contains $1$, so when you say #$G_m$, it counts $a$ as well).
There cannot be another $n' \in G.m$ that has a different number of ways. The same argument as above applies. If there is $a' \in G$ such that $a'm = n'$, then $a'G_mm = a'm = n'$. Note that there cannot be $x \in G, x \notin a'G_m$ such that $xm = n'$ because in that case, $$xm = n' \implies m = x^{-1}n' \implies a'G_mm = a'G_mx^{-1}n' \implies n' = a'G_mx^{-1}n' \implies a'G_mx^{-1} = 1 \implies x = a'G_m,$$ which is a contradiction.
This observation is the departure point for any and all applications of group actions. What you have is actually a bijection between $G/G_m$ and $G.m$, $aG_m \mapsto am$. This to me is one of the most remarkable expedients of group actions. When someone says a group is acting on a set, it immediately tells us a whole bunch of things about the group and the set. (When I read about group actions for the first time, I thought that one could, without much effort, define an action between any group and any set. This is NOT TRUE.)
The orbits of a group action are equivalence classes for the relation $x \sim x'$ if $x' = gx$ for some $g \in G$. The orbits partition the set, i.e. the set can be expressed as a disjoint union of orbits.
This image is what you must keep in mind when you know a group is acting on a set. 