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

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!

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 $$\Omega$$.