Consider a smooth action $G\curvearrowright M$ of a Lie group on a smooth manifold.
Suppose that an orbit $G\cdot p$ is closed. Is the orbit an embedded submanifold?
In general we know that the orbits are injective immersed submanifolds and if an orbit is embedded, then it is closed. Does the converse hold?
If the action is proper then the orbits are embedded, so the interesting question is about non proper actions.
Best Answer
The answer is positive even in greater generality:
Suppose that $G$ is a locally compact group with (at most) countably many connected components, $M$ a locally compact Hausdorff topological space, $G\times M\to M$ a continuous action. Let $p\in M$ be a point whose $G$-orbit $Gp$ is closed in $M$. Then the orbit map induces a homeomorphism $$ G/G_p \to Gp\subset M, $$ where $Gp$ is equipped with the subspace topology. Here $G_p$ is the stabilizer of $p$ in $G$. This result is Theorem 2.13 in the book by D.Montgomery and L.Zippin "Topological Transformation Groups''.
Note that Theorem 2.13 is stated in the case of transitive group actions. However, in general, the action of $G$ on its orbit is transitive and the assumption that the orbit is closed implies that the orbit is a locally compact space (since we are assuming that $M$ is).
Note that for this to apply you have to assume that Lie groups are, say, 2nd countable (some people do not make this assumption), so that they have (at most) countably many components. Otherwise, the result is false.