Let $A$ be a commutative ring and $I\subset A$ be an ideal.
Moreover, for a group $G$, we are given group action $G\curvearrowright A$ such that $g(I)=I$ for any $g\in G.$ Then, naturally $G$ acts on $A/I$.
Now, ideal $I^G:=\langle a \in I\mid g\cdot a=a (\forall g\in G)\rangle\subset A$ and $(A/I)^G:=\{a\in A/I\mid g\cdot a=a (\forall g\in G)\}$.
I want to know the Relation between $(A/I)^G$ and $A/I^G$.
Especially, the case of $A=K[X_1,\dots,X_n]$ and $I=(f_1,\dots,f_m)$, and $G$ is finite abelian group. Then $\operatorname{Spec}((A/I)^G)=V(f_1,\dots,f_m)/G\subset \mathbf{A}^n/G$.
On the other hand, $V(f_1,\dots,f_n)\subset\operatorname{Spec}(A/I^G)\subset \mathbf{A}^n$.
If we want to take a closed immersion $V(f_1,\dots,f_m)/G$ to $\mathbf{A}^n$, can we use $\operatorname{Spec}(A/I^G)$ well?