$\DeclareMathOperator{\Hom}{Hom}$ $\DeclareMathOperator{\k-alg}{k-alg}$
Let $A$ be a finitely generated algebra over a field $k$. Let $G$ be a finite group acting on $A$ as $k$-algebra automorphisms. Then $A^G = \{ a \in A : \sigma(a) = a \textrm{ for all } \sigma \in G\}$ is a finitely generated $k$-algebra, and $A$ is finitely generated as a module over $A^G$. Let $X = \operatorname{Spec} A$, $Y = \operatorname{Spec} A^G$, and $X \rightarrow Y$ the morphism corresponding to inclusion.
Is there a good description of the set of $k$-algebra homomorphisms $Y(k) = \Hom_{\operatorname{k-alg}}(A^G,k)$ in terms of $X(k) = \Hom_{\operatorname{k-alg}}(A,k)$?
When $k$ is algebraically closed, $X(k) \subset X$ and $Y(k) \subset Y$ are just the spaces of maximal ideals of $A$ and $A^G$ (this is just the Hilbert Nullstellensatz). The map $X \rightarrow Y$ is on the level of topological spaces a quotient map, which induces a quotient map $X(k) \rightarrow Y(k)$ on the spaces of closed points. Here $Y$ (resp. $Y(k)$) are the quotients of $X$ (resp. $X(k)$) under the action of $G$.
So here we can say that the $k$-algebra homomorphisms $A^G \rightarrow k$ are just the restrictions to $A^G$ of the $k$-algebra homomorphisms $A \rightarrow k$, where for $\phi, \phi': A \rightarrow k$, we have $\phi|A^G = \phi'|A^G$ if and only if $g.\phi = \phi'$ for some $g \in G$.
For $k$ not algebraically closed, I would expect there to be no such nice description. I am thinking of a similar situation for algebraic groups: if $N$ a normal subgroup of an algebraic group $H$, both defined over $k$, the failure of the injection $H/N(k) \rightarrow H(k)/N(k)$ to be an isomorphism is measured by the size of the cohomology set $H^1(\operatorname{Gal}(k_s/k), N(k_s))$. Is there a similar cohomological invariant which can describe the situation here?