One has to be slightly careful here regarding the notion of smoothness, particularly the smooth structure on the orbit of a point. Suppose a Lie group $G$ acts smoothly (not necessarily freely) on a smooth manifold $M$, and let’s fix $p\in M$. Define the map $\theta_p:G\to M$, $g\mapsto g\cdot p$.
$\theta_p$ as a mapping of $G$ into $M$ is certainly smooth, since the group action is.
For every $g_1,g_2\in G$, we have $\theta_p(g_1g_2):=(g_1g_2)\cdot p=g_1\cdot (g_2\cdot p)=g_1\cdot \theta_p(g_2)$, i.e the mapping $\theta_p$ is $G$-equivariant relative to the natural transitive left action of $G$ on itself, and the given left action of $G$ on $M$. Hence $\theta_p$ has constant rank.
The stabilizer $G_p$ is an embedded Lie subgroup of $G$ (it is embedded since it is the $p$-level set of the constant rank map $\theta_p$; in fact $G_p$ has codimension equal to the rank of $\theta_p$). Equipping $G/G_p$ (the quotient of $G$ by the right-translation action of $G_p$) with the quotient topology, the quotient manifold theorem tells us there’s a unique smooth structure which makes $\pi:G\to G/G_p$ into a smooth surjective submersion.
By definition of the stabilizer $G_p$, the map $\theta_p$ is constant on the fibers of $\pi:G\to G_p$, so it descends to the quotient, i.e there is a unique mapping $f_p:G/G_p\to M$ such that $\theta_p=f_p\circ \pi$ (i.e a certain triangular diagram commutes). Since $\pi$ is a surjective submersion, it follows (this is known as the universal property of surjective submersions) that $f_p$ is smooth as well and in fact has the same constant rank as $\theta_p$. Finally, the mapping $f_p$ is injective (almost by definition since we quotiented out the stabilizer), so the global rank theorem implies that $f_p$ is an injective immersion.
Notice carefully I only claimed that the map $f_p:G/G_p\to M$ is smooth (and an injective immersion). I cannot arbitrarily shrink the target to its image, $\mathcal{O}_p:=G\cdot p$ (the orbit), and claim the restricted map $\widetilde{f_p}:G/G_p\to \mathcal{O}_p$ is also smooth. To make such statements we have to be a little careful about the smooth structure on the orbit $\mathcal{O}_p$. Here is what we can say: since $\widetilde{f_p}$ is a bijection, we can transport the topology and smooth structure of $G/G_p$ onto $\mathcal{O}_p$ so it becomes a diffeomorphism, and with this structure, the inclusion $\iota_p:\mathcal{O}_p\to M$ factorizes as $\iota_p=f_p\circ (\widetilde{f_p})^{-1}$ which is a composition of an injective immersion with a diffeomorphism, i.e is an injective immersion, so $\mathcal{O}_p$ is an immersed submanifold of $M$. So, to interpret the diffeomorphism $G/G_p\cong \mathcal{O}_p$, we have to be careful to note that $\mathcal{O}_p$ is not necessarily an embedded submanifold of $M$.
In some special cases, the orbit $\mathcal{O}_p$ is actually an embedded submanifold and the target-restricted map $\widetilde{f_p}:G/G_p\to \mathcal{O}_p$ is smooth and a diffeomorphism. In increasing order of speciality, this occurs if:
- the orbit $\mathcal{O}_p$ is locally closed in $M$ (proving this is sufficient uses Baire’s category theorem)
- the orbit $\mathcal{O}_p$ is closed in $M$
- the orbit relation $\mathcal{R}:=\{(x,y)\in M\times M\,: \text{$x$ and $y$ lie in the same orbit}\}$ is closed in $M\times M$ (note that since the projection $M\to M/G$, as topological spaces, is an open quotient map, closedness of $\mathcal{R}$ is equivalent to $M/G$ being Hausdorff)
- $G$ acts properly on $M$ (i.e the map $G\times M\to M\times M$, $(g,x)\mapsto (g\cdot x,x)$ is a proper map, meaning preimages of compact sets are compact)
In the case of a free action, the stabilizer is trivial, so we don’t really need to pass to the quotient, but still, to argue that the orbit $\mathcal{O}_p$ is immersed, one needs an argument such as the one above, and making some extra assumptions, one gets the embedded case. So in either case, we have a diffeomorphism $G\cong \mathcal{O}_p$, but the diffeomorphism has to be interpreted suitably.
Finally to show that these distinctions are actually necessary, consider the $\Bbb{Z}$-action on $S^1$ which is translation by an irrational argument; then each orbit is dense. Another example to keep in mind is the dense curve on a torus
