@pastebee The quantifier elimination that occurs is in an expanded language, where the additional definitions do use quantifiers. So I don't think it quite gets what we want here.

@JamesEHanson Why do you say it's probably bi-interpretable with it? $\mathbb{R}$ has an ordering after all ...

Or if you prefer an existential second-order definition, the set of numbers which are fixed by some nontrivial automorphism.

@JamesEHanson Ooh, I didn't know that - that's neat (although I don't think it immediately helps here)!

@JoelDavidHamkins Ah, I see. I don't see how to make that work immediately though. For example, it could be (to the best of my knowledge) that $F$ has two nontrivial automorphisms $f,g$ each of finite order such that the sequence $fg,fgfg,fgfgfg,...$ has no limit in that topology.

@Wojowu The set of numbers not moved by any automorphism. Under AD, the only automorphism is conjugation.

@cody Yes, that's correct.

@JoelDavidHamkins That's the issue I was getting at with my above comment. I don't see that an $F$ need have a good topology (in $L(\mathbb{R})$) at all, so I don't know how to take limits.

@JoelDavidHamkins I thought a bit along those lines but didn't get anywhere. One issue for me is that the doppelganger $F$ might, within $L(\mathbb{R})$, look too large for descriptive set theoretic techniques (at least the ones I'm familiar with) to apply. E.g. if $\vert F\vert^{L(\mathbb{R})}>\Theta$, which I don't immediately see is impossible, I don't see how to do anything useful.

@AsafKaragila I agree, which is why I asked the question - I'm hoping for a negative answer.

@JoelDavidHamkins I've added a follow-up question you might find interesting.

@JoelDavidHamkins Thanks! The intuition is basically the same as in this old answer of mine.