How many elementary equivalent models are unifiable by ultrapower? Definition. A class $\mathcal{C}$ of pairwise elementary equivalent $\mathcal{L}$-structures is unifiable by ultrapower if there is an index set $I$ and an ultrafilter $F$ on it such that $\forall M,N\in \mathcal{C}~~~\prod_{F}M\cong \prod_{F}N$.
A theorem by Shelah and Keisler says that each set of elementary equivalent structures with size $2$ is unifiable by ultrapower.
Question. For which cardinal $\kappa$ the following assertion is true?
Each set $\mathcal{C}$ of size $\kappa$ of pairwise elementary equivalent $\mathcal{L}$-structures is unifiable by ultrapower. 
What about when $\mathcal{C}$ is a proper class of pairwise elementary equivalent $\mathcal{L}$-structures?
The question is closely related to the problem of iterated ultrapowers.
 A: A class $C$ of elementarily equivalent structures is unifiable by ultrapower if and only if the structures in $C$ have bounded cardinality (in other words, if they have a set of representatives up to isomorphism).
The left-to-right implication was given in Noah’s comment above: if we fix $M\in C$, then every $N\in C$ satisfies
$$|N|\le\Bigl|\prod_FN\Bigr|=\Bigl|\prod_FM\Bigr|=:\kappa.$$
As for the converse implication, Shelah proved in Every two elementarily equivalent models have isomorphic ultrapowers something considerably stronger than just the Shelah–Keisler theorem:

Theorem (Shelah). Let $\lambda$ be an infinite cardinal, and $\mu=\min\{\mu:\lambda^\mu>\lambda\}$. There exists an ultrafilter $F$ on $\lambda$ with the following property.
Whenever $\{M_\alpha:\alpha<\lambda\}$ and $\{N_\alpha:\alpha<\lambda\}$ are sequences of models of cardinality $\le\kappa<\mu$ in the same language such that the ultraproducts $\prod_{\alpha<\lambda}M_\alpha/F$ and $\prod_{\alpha<\lambda}N_\alpha/F$ are elementarily equivalent, then they are isomorphic.
In particular, if $M,N$ are elementarily equivalent models of cardinality $<\mu$, then the ultrapowers $M^\lambda/F$ and $N^\lambda/F$ are isomorphic.

Note that for a given $\kappa$, we can take e.g. $\lambda=2^\kappa$ to make $\mu>\kappa$.
