Let $L$ be the ground model, and $a\in2^\omega$ be a Sacks-generic real over $L$. Note that any real $x\in S=(2^\omega\cap L[a])\setminus L$ is still Sacks-generic over $L$. Now assume that $\mathsf E$ is an OD (ordinal-definable) equivalence relation in $L[a]$ on the set $S$, with exactly two equivalence classes, say $M$ and $F$. (Two genders of the Sacks reals.) Are $M$ and $F$ necessarily OD themselves?
My idea of a counterexample is as follows. Let $F$ be the set of all continuous 1-1 maps $2^\omega$ onto $2^\omega$, coded in $L$. Then $F$ is a group under the superposition. Moreover if $x,y\in S$ then it is known that $y=f(x)$ for some $f\in F$. Now if $H\subseteq F$ is a subgroup coded in $L$ then the relation:
$x \mathrel{\mathsf E_H} y$ iff $y=f(x)$ for some $f\in H$
is an OD equivalence on $S$, and there is no immediate idea as how to OD-define the $\mathrel{\mathsf E_H}$-class of $a$ (w/o a reference to $a$). Now the goal is to define $H$ such that $\mathrel{\mathsf E_H}$ has exactly two equivalence classes on $S$ in $L[a]$. The principal non-commutativity of $F$ looks to be a huge obstacle though.
[Added Feb 4 at 5:50] And finally it is established, there are two distinct but OD-indiscernible populations of Sacks-generic reals over $L$, arXiv . In fact this is an unpublished result of Solovay dated back to 2002. A similar result holds for $E_0$-large forcing, via a known ``canonization'' theorem, but the problem is open for other popular forcing notions.
