This question is partly motivated by Timothy Chow's recent question on the division paradox.

Say that a set $X$ admits a *paradoxical partition* if and only if there is an equivalence relation $\sim$ on $X$ such that $|X|<|X/{\sim}|$, that is, there is a partition $$ X=\bigcup_{i\in I}X_i $$ of $X$ into disjoint, non-empty sets $X_i$, $i\in I$, for some set $I$ with $|X|<|I|$ (we split $X$ into more pieces than elements). No such partitions are possible if $X$ is finite or, more generally, well-orderable but, in the absence of choice, examples can be found. Famously (as mentioned in Chow's question), in Solovay's model we have $|\mathbb R|<|\mathbb R/E_0|$, where $E_0$ is Vitali's equivalence relation that identifies two reals precisely when their difference is rational. The same example holds under determinacy.

Assume determinacy and work in $L(\mathbb R)$. Suppose that $X$ is not well-orderable (this means that $\mathbb R$ injects into $X$). Does $X$ admit a paradoxical partition?

In comments, Gabe Goldberg suggested that this may be the case, which I also suspect, but I do not see how to proceed. In particular, I do not see how to take advantage of a paradoxical partition of $\mathbb R$ to obtain one of $\mathcal P(\mathbb R)$. (Note that $2^{|\mathbb R|}=2^{|\mathbb R/E_0|}$.)

Working in $L(\mathbb R)$ under determinacy is more or less just a test case, and any suggestions or partial answers under other assumptions are definitely welcome.