I saw the below content in Tao's blogs. I don't know why the manifold equal to the quotient of the group by the stabiliser?
Besides, the stabiliser depend on $x$, according to Wiki. For example, $G$ is a group, $x\in G$, then $$G_x=\{g\in G : gx=x\}$$ is a stabiliser of $x$. Although it is not different in $S^3$. But it is different in other manifold.

It is a fact from group theory known as orbit-stabilizer theorem. To understand let's define the needed quantities:
Group action: An action of a Group $G$ on a set $M$ is a map $$\begin{cases} G\times M\to M\\ (g,x)\to gx \end{cases}$$ which satisfies $g_1(g_2x)=(g_1g_2)x$ for $g_1,g_2\in G$ and $x\in M$
Stabilizer: The stabilizer of a point with regards to a group action is defined as the set $$G_x:=\{g\in G|gx=x\}$$
Orbit: The orbit of a point with respect of the group action is defined as
$$Gx:=\{gx|g\in G\}$$
We find using this we can define an equivalence relation $x\sim y:\Leftrightarrow y\in Gx$ using which we can define the quotient space $G/G_x$ which leads to the promised theorem:
Theorem1: The map $\begin{cases} G/G_x\to Gx \\ gG_x\mapsto gx\end{cases}$ is an bijection.
If we now have a transitive action this in paricular means that $Gx=M$. If we now want to adapt this to the setting of smooth manifolds we have to add a further theorem:
Theorem2: Let $G$ be a Lie group and $H\le G$ a closed subgroup. Than there is a unique smooth structure on $G/H$ and the quotient map $G\to G/H$ is a smooth fibration.
Now this means that we have a smooth structure on $G/H$ and $M$ and that the two spaces are connected by a bijection. If we further add that the action of $G$ on $M$ is a smooth map we get a diffeomorphism between the two smooth manifolds.