Is the fixed point set of an action a submanifold?

The fixed-point set can be extremely wild. For example, every closed subset of $\mathbb R^n$ is the fixed point set of some smooth $\mathbb R$-action. To see this, let $S\subseteq\mathbb R^n$ be closed, and let $f\colon \mathbb R^n\to\mathbb R$ be a smooth nonnegative function whose zero set is exactly $S$. (Such a function always exists; see, for example, Theorem 2.29 in my Introduction to Smooth Manifolds, 2nd ed.) Now let $V$ be the following vector field on $\mathbb R^n$: $$ V_x = \frac{f(x)}{1+f(x)} \,\left.\frac{\partial}{\partial x^1}\right|_x. $$ The fact that the coefficient of $V$ is bounded in absolute value by $1$ guarantees that $V$ is complete (exercise for the reader), so its flow is a smooth $\mathbb R$-action on $\mathbb R^n$. The fixed points of this action are exactly the zeros of $V$, which are the points of $S$.

Let $x \in M$ be a fixed point of a $G$-action, $G$ acting properly on $M$ (meaning the map $G \times M \to M \times M$, $(g,m) \mapsto (m,gm)$, is proper). We'll use the following tubular neighborhood theorem:

If $M$ is a smooth proper $G$-manifold without boundary, and $N$ is a closed $G$-invariant submanifold, then there is an open neighborhood $U \supset N$ which is $G$-equivariantly diffeomorphic to the total space of a $G$-vector bundle over $N$ (that is, a vector bundle over $N$ such that the action of $G$ on $N$ extends to a fiberwise linear action of $G$ on the bundle.) See proposition 4.5 here.

Now, let's kill the fly. Since $x$ is a fixed point, it's a $G$-invariant submanifold. A $G$-vector bundle over it is the same thing as a vector space with a linear action of $G$. The fixed point set of this action is then necessarily a vector subspace. Transporting this back to the neighborhood of $x$, we see that locally near $x$, the fixed point set forms a submanifold, as desired.

(Indeed, the set of points $x$ such that the isotropy group of the action at $x$ contains some fixed closed subgroup $H \subset G$ forms a submanifold by the exact same argument; of course, the example above corresponds to the case $H=G$.)