Let $R$ be a commutative algebra over $\mathbb{C}$. Assume $R$ is noetherian. Let $G$ be a finite group acting on $R$ with the action on $\mathbb{C}$ trivial. Let $M$ be a $G$-equivariant $R$ module, i.e. there is a $G$-action on $M$ such that $$ g(rm)=(gr)(gm). $$
Now assume $M$ is a finitely generated $R$-module, i.e. there exists a surjective $R$-module homomorphism $$ \phi: R^{n}\twoheadrightarrow M. $$ Here we do not assume that $\phi$ is $G$-equivariant.
My question is: is $M$ also $G$-equivariantly finitely generated, i.e. can we find a $G$-equivariant surjective $R$-module homomorphism $$ \tilde{\phi}: R^{\tilde{n}}\twoheadrightarrow M? $$
No, although this has nothing to do with finiteness--the problem is that the images of all the equivariant homomorphisms $R\to M$ may not generate $M$. To see this, note $1\in R$ is fixed by every element of $G$, so the image of any equivariant homomorphism $R\to M$ is generated by an element that is fixed by $G$. But the elements of $M$ that are fixed by $G$ may not generate $M$. For instance, let $R=\mathbb{C}[x,y]$ with $G=S_2$ acting by swapping $x$ and $y$, and let $M=R/(x,y)$ with the nontrivial element of $G$ acting by multiplication by $-1$. Then the only element of $M$ that is fixed by $G$ is $0$.