Let $G$ be a Lie group and $H$ a Lie subgroup of $G$.
Let $M$ be a smooth manifold.
Let $\theta$ be a left smooth action of $G$ on $M$.
Let $S=\{p\in M| G_p=H\}$, where $G_p$ is the isotropy group of $p$.
Is $S$ a smooth submanifold of $M$?
Let $G$ be a Lie group and $H$ a Lie subgroup of $G$.
Let $M$ be a smooth manifold.
Let $\theta$ be a left smooth action of $G$ on $M$.
Let $S=\{p\in M| G_p=H\}$, where $G_p$ is the isotropy group of $p$.
Is $S$ a smooth submanifold of $M$?
More details in Thomas Rot's comment. Every closed subset of any manifold is the set of zeroes of a vector field, even one which is complete (for example, bounded in a complete Riemannian metric). So the vector field generates a group action, whose fixed point set is that closed set. So if we take $G$ to be the real line, and $H=G$, we get any closed set arising as $S$, for some vector field. To be more explicit, if $f(x)$ is bounded and vanishes to all orders at the origin, and nowhere else, then $\sin(1/x)f(x)\partial_x$ can be our vector field on the real line, with fixed points $x$ at $1/x \in 2\pi \mathbb{Z}$.