Bijection of 2-colorings and permutation equivalence classes Let $G$ be a group of permutations of  $n$ objects. Let subsets $A$,$B$ of the objects be equivalent if $A=g(B)$ for some $g \in G$. 
Is there a way to show that a bijection exists between the equivalence classes and the $2$-colorings of the objects that are invariant under those permutations in $G$? 
So Far, I have verified that it works(their cardinalities are the same) for a few cases I tried, but I do not see how it can be generalized. 
I will post an example, the definition should be clear from the example. 
let $G=\{e,(2 3)\}$, $n=3$
then, we have 8 subsets and 6 equivalent subsets. namely $\{\}, \{1\},\{\{2\},\{3\}\}, \{\{1,2\},\{1,3\}\},\{2,3\},\{1,2,3\}$ 
and the invarient transformations are. 
$\{1,1,1\},\{1,1,0\} \sim \{1,0,1\},\{0,1,1\},\{1,0,0\},\{0,0,1\} \sim \{0,1,0\},\{0,0,0\}$
In this example applying (2 3) to {0,1,0} changes it to {0,0,1} and applying e, leaves it as it is. 
 A: So, what you are actually doing is this: 
Let $X$ be a set, and let $G$ be a subgroup of $S_X$, the group of permutations on $X$.
We define two equivalence relations, one, let's call it $\sim$, on $\mathcal{P}(X)$, the power set of $X$, and one, let's call it $\equiv$, on the set of functions $2^X = \{f\colon X\to \{0,1\}\}$, which we think of as the $2$-colorings of $X$. We define:
\begin{align*}
A\sim B &\Longleftrightarrow \text{there exists $\sigma\in G$ such that $\sigma(A)=B$}\\
f\equiv g &\Longleftrightarrow \text{there exists $\sigma\in G$ such that $f\circ\sigma = g$.}
\end{align*}
The question is whether there is a bijection between $\mathcal{P}(X)/\sim$ and $2^X/\equiv$.
The key is that there is a natural identification of $2^X$ with $\mathcal{P}(X)$, by mapping $f\colon X\to\{0,1\}$ to the subset $f^{-1}(1) = \{x\in X\mid f(x)=1\}$, and mapping $A\subseteq X$ to $\chi_A$, the characteristic function of $A$.
Define $\mathcal{F}\colon \mathcal{P}(X)/\sim \to 2^X/\equiv$ as follows: given a class $[A] = \{ B\in\mathcal{P}(X)\mid A\sim B\}$, we let
$$\mathcal{F}([A]) = \{ \chi_B\mid B\in [A]\}.$$
First, I claim this is well defined: that is, I claim that $\mathcal{F}([A])$ is an equivalence class modulo $\equiv$.
Suppose $A\sim B$. Then there exists $\sigma\in G$ such that $\sigma(A)=B$. I claim that $\chi_B\circ\sigma = \chi_A$. Indeed, let $x\in X$. If $x\in A$, then $\sigma(x)\in B$, so $\chi_B\circ\sigma(x) = 1 = \chi_A(x)$. And if $x\notin A$, then $\sigma(x)\notin B$ (since $A=\sigma^{-1}(B)$), so $\chi_B\sigma(x) = 0 = \chi_A(x)$. Thus, $\chi_B\equiv \chi_A$. 
Conversely, suppose that $f\in 2^X$ is equivalent to $\chi_A$. Then there exists $\tau\in G$ such that $f\circ\tau = \chi_A$. Let $B=\tau(A)$. I claim that $\chi_B=f$. Indeed, let $x\in X$. Then $f(\tau(x))=1$ if and only if $x\in A$, if and only if $\tau(x)\in B$. So $f(y)=1$ if and only if $y\in B$. Note that $B\sim A$.
Therefore, every element of $[A]$ gets mapped to an element of
$$\langle \chi_A \rangle = \{ f\in 2^X\mid f\equiv \chi_A\}$$
and every element of $\langle\chi_A\rangle$ comes from an element of $[A]$. Thus, $\mathcal{F}$ is well defined.
Now, $\mathcal{F}$ is one-to-one: if $\mathcal{F}([A]) = \mathcal{F}([B])$, then 
$$\chi_B\in \langle \chi_B\rangle = \mathcal{F}([B]) = \mathcal{F}([A])=\langle \chi_A\rangle$$
so there exists $\sigma\in G$ such that $\chi_B\sigma = \chi_A$. that means that $x\in A$ if and only if $\sigma(x)\in B$, so $B=\sigma(A)$, hence $B\sim A$, so $[A]=[B]$.
And $\mathcal{F}$ is onto: let $\langle f\rangle$ be an equivalence class of $2^X/\equiv$. Let $A = f^{-1}(1)$. I claim that $\mathcal{F}([A])=\langle \chi_A\rangle = \langle f\rangle$. Indeed, $\chi_A = f$ by construction, so $f\in\langle\chi_A\rangle$, hence $\langle f\rangle = \langle \chi_A\rangle = \mathcal{F}([A])$.
Therefore, $\mathcal{F}$ is a bijection between $\mathcal{P}(X)/\sim$ and $2^X/\equiv$. 

However, I want to add that I think it is fair to say that your nomenclature is misleading/nonstandard. I would take that a $2$-coloring of $X$ is invariant under $G$ to mean that if $f\colon X\to\{0,1\}$ is the $2$-coloring, then $f\circ \sigma = f$ for all $\sigma\in G$. 
A: I think this will work. 
Consider the bijection where coloring $\{...,x_i=1,...\}$ means that object $i$ is included in our set, then the rest follows. 
