Let $G$ be a Polish group, $X$ a Polsih space on which $G$ acts continously and consider the orbit equivalence relation on $X$ with respect to $G$.
Suppose $A\subseteq X$ is a perfect set of pairwise non orbit equivalent elements, $B\subseteq X$ is closed and for all $a\in A$, the $G$ orbit of $a$ contains exactly one $G$ orbit of $B$.
My question: Does it follow that $B$ contains a perfect set of pairwise non orbit equivalent elements?
I assume the answer is positive but I cannot prove it. One idea I have been entertaining is to show that for some $g\in G$, there are uncountably many $a\in A$ such that $g\circ a\in B$, where $\circ$ denotes the action of $G$ on $X$. Clearly, if this can be done, then by the perfect set theorem the question is answered.
Any help is appreciated.
Sorry about my earlier false attempt. I think this one now is slightly better.
Consider the Polish space $A\times G$ with a subset $S$, such that $(a,g)\in S$ iff $g\circ a\in B$. Clearly $S$ is closed, and $\operatorname{proj}_A(S)=A$ by assumption. By Jankov-von-Neumann uniformization, there is a $\sigma(\mathbf{\Sigma^1_1})$ measurable function $f:A\to G$ such that $f\subseteq S$. In particular, $f$ is Baire mesurable. Recall that every Baire measurable function is continuous on a comeager set, i.e. we may fix some comeager $G_\delta$ set $C\subseteq A$ such that $f\upharpoonright C$ is continuous. Now because that for any $a\ne a'\in A$ we have $f(a)\circ a\nsim_G f(a')\circ a'$, the set $T=\{f(a)\circ a\mid a\in C\}\subseteq B$ is $G$-independent. As $C$ is comeager $G_\delta$, the set $T$ is also Borel uncountable. By the perfect set property of Borel sets, $T$ has a perfect $G$-independent subset as desired.