This is yet another question about cardinalities in forcing extensions of models of $\mathsf{ZF+\neg AC}$ (see also here). Specificially, I think I've isolated the simplest question which I can't yet answer. I suspect this is actually quite easy and I'm just having a silly moment, but currently I don't see the argument:
Suppose $V$ is a transitive model of $\mathsf{ZF+AD}$. Let $c,d$ be mutually Cohen generic over $V$; is there in $V[c,d]$ a bijection between $\mathbb{R}^{V[c]}$ and $\mathbb{R}^{V[d]}$?
Note that a negative answer to the above-linked question in the case of Cohen forcing would give a positive answer to this question.
"Obviously" the answer should be yes, but I don't see how to prove that. We have a set $\mathcal{X}\in V$ of names for reals such that every real in a Cohen extension is named by some element of $\mathcal{X}$, and each Cohen real $a$ induces an equivalence relation $\sim_a$ on $\mathcal{X}$ as $\nu\sim_a\mu\leftrightarrow\nu[a]=\mu[a]$. So really this question is asking for a bijection in $V[c,d]$ between $\mathcal{X}/\sim_c$ and $\mathcal{X}/\sim_d$. Intuitively this should exist since Cohen forcing is as homogeneous as one could hope; however, in the absence of choice in $V[c,d]$ I don't actually see how to build one.
EDIT: While it's not my main question, I'd also be interested in a weak negative result: are there $M\models\mathsf{ZF}$ and mutually-Cohen-over-$M$ reals $c,d$ such that $M[c,d]\models \mathbb{R}^{M[c]}\not\equiv\mathbb{R}^{M[d]}$? I am really interested in the determinacy case, but any sort of pathology like this would be quite cool.
To preempt one natural attempt, note that Cohen forcing kills determinacy so we can't use determinacy in $V[c,d]$ even though we have it in $V$. While at first glance this might appear to contradict (say) the generic absoluteness of the theory of $L(\mathbb{R})$ given large cardinals, there is no discrepancy since $(L(\mathbb{R}))^V[G]\not=(L(\mathbb{R}))^{V[G]}$ in general.
To move this off the unanswered queue, I'll observe that Paul Larson gave a negative answer at MO. (I've made this CW to avoid reputation gain.)