Suppose $T$ is a complete first-order theory (in an finite, or at worst countable, language). Given any model $\mathcal{M}\models T$ of cardinality $\kappa$, we can ask whether $\mathcal{M}$ can be made isomorphic to any countable model $\mathcal{N}\models T$ in the ground model; that is, whether there is some $\mathcal{N}\models T$, which is countable, such that $V[G]\models \mathcal{M}\cong\mathcal{N}$ for $G$ $Col(\kappa,\omega)$-generic over $V$. **For simplicity, we'll say "$\mathcal{M}$ collapses to $\mathcal{N}$".**

In case $T$ has *countably many* countable models, the answer is yes: the set of models of $T$ is a countable Borel set, and so is absolute. The uncountable models of such a $T$ are thus partitioned into $\omega$-many classes. My question is about what the nonempty classes are. For example, suppose $T$ is $\aleph_1$-categorical. Then the countable models of $T$ form an elementary chain, with a "top" model $\mathcal{N}_\omega$, and it is easy to see that this is the only countable model which uncountable models can collapse to.

However, there are plenty of non-$\aleph_1$-categorical theories with only countably many countable models, and beyond the $\aleph_1$-categorical setting things are much less clear. Some examples:

The language has predicates $P_i$ ($i\in\omega$); $T$ asserts that the $P_i$ name infinite disjoint sets. Then the countable models of $T$ are classified by how many elements they have not in $\bigcup_{i\in\omega} P_i$, and so there are countably many countable models. For each countable model of $T$, there is an uncountable model of $T$ which collapses to it.

The language has a single binary relation $E$; $T$ asserts that $E$ is an equivalence relation and that there is exactly one class of cardinality $n$, for each $n$. Countable models of $T$ are determined by how many infinite classes they have; and every countable model of $T$

**except the prime**is collapsed to by some uncountable model of $T$.

I'm curious what we can say in general about the set of models which can be collapsed to. There's a lot of questions around here that one can ask; let me focus on:

Is there a theory $T$ with countably many countable models, including $\mathcal{M}_0\prec\mathcal{M}_1$, such that $\mathcal{M}_0$ can be collapsed to by some uncountable structure but $\mathcal{M}_1$ cannot?

That is, can there be "gaps" in the range of the collapse?

EDIT: As Paul Larson pointed out in the comments below, a countable model is collapsed to by some uncountable model iff it is **extendible** (see Definition 2.6 in http://shelah.logic.at/files/1003.pdf) - that is, iff it has an uncountable $L_{\omega_1\omega}$-elementary extension. One direction is trivial. To show that every extendible model is collapsed to, let $\mathcal{M}$ be extendible with $\mathcal{M}\prec_{\omega_1\omega}\mathcal{N}$ for $\mathcal{N}$ uncountable. Let $\varphi$ be the Scott sentence of $\mathcal{M}$, and let $V[G]$ be a forcing extension in which $\mathcal{N}$ is made countable. Then $\mathcal{N}\models\varphi$ in both $V$ and $V[G]$; moreover, the statement "$\varphi$ is the Scott sentence of $\mathcal{M}$" is $\Pi^1_2$ (actually, better, but this is enough) and so absolute between $V$ and $V[G]$. So $V[G]\models\mathcal{M}\cong\mathcal{N}$.

It's worth keeping in mind the broader result, due to Barwise and Karp if I recall correctly, that $\mathcal{M}\equiv_{\infty\omega}\mathcal{N}$ iff $\mathcal{M}\cong\mathcal{N}$ in some forcing extension.