I wanna show that $(K^X)^G \simeq K^Y$. Well, I will explain what is $X,Y$ and $G$ in this context.
Let $X$ be a nonempty set and $G$ a group, with an action over $X$. Let $K^X$ be the algebra of functions from $X$ to $K$, where $K$ is a field and consider the translation $\tau: X \times G \to X \times X$ given by $(x,g) \mapsto (x,x \cdot g)$.
If $Y = X/G$(induced by $\tau$) and $(K^X)^G$ are elements of $K^X$ stable under $G$, then $(K^X)^G \simeq K^Y$.
Honestly, I don't know how to start that. So I don't wanna an complete answer neither a flag. I just ask you some hint how to start, maybe define the morphism. So please, help me.