Do invariant functions form a Banach (sub)manifold in function spaces?

Question

Let $G$ be a topological group, and $X$ some function space; preferably a Sobolev space $X=W^{1,p}(\Omega)$, where $\Omega \subset \mathbb{R}^n$ is some invariant subset ($g\Omega \subset \Omega$) or the whole space.
Let $G$ be represented as linear, continuous operators $\pi(g)$ on $X$ in the form $$ \pi(g)f(x)=f(gx) $$
Does the set of invariant functions that satisfy $$ f(gx)=f(x) $$ form a Banach (sub)manifold of $X$?
Do we need additional assumptions on the group (e.g.compact groups)? As an example consider $G=O(n)$ the orthogonal group. Does the set of radial symmetric functions form a Banach (sub)manifold of $X$?
Feel free to modify and add assumptions if necessary as I am not too familiar with this field. I am thankful for any reference, hint or remark!

Answer 1

In most cases they form a closed subspace: it is subspace, because if $f$ and $g$ are invariant, then so is $f+g$ or $\lambda f$. Whenever it is closed or not depends on norm, but in case of $\sup$-norm or $L^2$ norm it is.

The subspace of such functions can be considered as functions on space of orbits $\Omega/G\Omega$, where measure induced from $\Omega$.