Question: Let $G,H$ be profinite groups of cardinality $|\mathbb{R}|$, with the same finite quotients (here I only consider quotients by normal, open subgroups). Then are $G$ and $H$ isomorphic?
Background: First, note that one needs some sort of cardinality assumption, as otherwise one could take $G=\mathbb{F}_2^I$ and $H=\mathbb{F}_2^J$ where $I$ and $J$ are infinite of distinct cardinality. Moreover, it is NOT sufficient to just require that $G$ and $H$ have the same cardinality, as one could take $G=\mathbb{F}_2^I\times \mathbb{F}_3^J$ and $H=\mathbb{F}_2^J\times\mathbb{F}_3^I$.
As for motivation, if we take $C$ to be an affine hyperbolic curve over the algebraic closure of a finite field, then it is known which subgroups occur as finite quotients of the etale fundamental group. Given this phrasing, I assume that this is not enough information to ecover the actual etale fundamental group, and so I wanted an explicit counterexample.
EDIT: Due to the comments, I think a better question is as follows: Rather than assuming $G$ and $H$ have cardinality $\mathbb{R}$, let me assume instead that they have at most countably many homomorphisms to any finite group.
This avoids (at least) some counterexamples assuming Luzin' hypothesis.
Let me add that the answer is yes if $G$ and $H$ are topologically finitely generated. This is well know, but as its quick I include a proof:
Note that for each positive integer $n$, $G$ has only finitely many homomorphisms to fintie groups of order at most $n$. Let $G(n)$ be the kernel of all of these, and $G_n$ the corresponding finite quotient. Then $G$ is the inverse limit of the $G_n$. To see this note that the natural map $G\to\varprojlim_n G_n$ is clearly injective, and as it is surjective on every quotient it must also be surjective by compactness.
So it is enough to see that $G_n\cong H_n$. But by assumption, $G$ surjects onto $H_n$, and such a surjection must factor through $G_n$, so $G_n$ surjects onto $H_n$. Likewise $H_n$ surjects onto $G_n$ and since these are finite groups they must be isomorphic.