This question was previously asked and bountied at MSE. Throughout, "theory" means "possibly-incomplete first-order theory in a countable language."
Say that a theory $T$ has CHS (= continuum hypothesis for substructures) iff for every countable $\mathcal{M}\models T$, the set of isomorphism types of substructures of $\mathcal{M}$ also satisfying $T$ is either countable or size continuum. For example, the theory of linear orders has CHS: since $\mathbb{Q}$ has continuum-many nonisomorphic suborders only scattered linear orders could possibly be counterexamples, and they're handled by Laver's inductive analysis of scattered linear orders (see the introduction of Equimorphism Invariants for Scattered Linear Orderings by Montalban).
My question is whether this is in fact trivial:
Assume Vaught's conjecture holds. Does every theory have CHS?
I don't see how to produce a counterexample, but I also don't see how to even start a proof of a positive result here.
Two quick comments:
There is no obvious descriptive-set-theoretic obstacle: given a countable structure $\mathfrak{A}$ we can think of substructures of $\mathfrak{A}$ as points in Cantor space, and isomorphism then corresponds to a $\Sigma^1_1$ equivalence relation on Cantor space, but $\Sigma^1_1$ equivalence relations on Cantor space can have $\aleph_1$-many classes even if $2^{\aleph_0}\not=\aleph_1$.
If $2^{\aleph_0}=\aleph_1$, then the question as written has a trivial affirmative answer. Analogously to the situation for Vaught's conjecture, there is an "absolute" version of the question: namely, ask whether every theory $T$ has the property that for every $\mathfrak{A}\models T$ with underlying set $\mathbb{N}$, there is a perfect set $P\subseteq 2^\omega$ such that the substructures of $\mathfrak{A}$ corresponding to the elements of $P$ are non-isomorphic. This is not trivialized by cardinal arithmetic, and in the end is probably the "right" version of the question to ask.
