Let $(\Omega, \mathcal{F}, P)$ be a probability space and $G$ a finite group of measurable, bijective maps $g: \Omega \rightarrow \Omega$ which are $P$ invariant, i.e. they have the property $P(g^{-1}(A)) = P(A) \quad \forall A \in \mathcal{F}$, and define $$\mathcal{C}_G \equiv \{A \in \mathcal{F}: g(A) = A \quad \forall g \in G\}$$
It is trivially seen that $\mathcal{C}_G$ is a $\sigma-$algebra. I want to show the following: $$\textbf{(I)} \quad \quad E(X | \mathcal{C}_G) = \frac{1}{|G|}\sum_{g \in G} X \circ g(\omega) \quad P-\text{a.s.}$$
My attempt: Define $Y$ as the RHS of (I). It is easily shown that $E(Y1_A) = E(X1_A) \quad \forall A \in \mathcal{C}_G$ so it suffices to show that the RHS is $\mathcal{C}_G$ measurable. And for this we simply need to show that $X \circ g$ is $\mathcal{C}_G$ measurable for each $g \in G$. I don't really know how to do this, because what I need to prove is the following:
Fix $g \in G$ and then show $\forall h \in G$,
$$h(g^{-1}(X^{-1}(B))) = g^{-1}(X^{-1}(B))$$
Does anyone have any ideas? Any help would be massively appreciated. Thanks!
Hint: For any $g_1 \in G$ we have $Y\circ g_1 =Y [P] \, a.s.$ since $\{gg_1: g \in G\}=G$. This implies that $Y$ is measurable w.r.t the $P-$ completion of $\mathcal C_G$.