Skip to main content
5 of 6
simplified example

It is a fact (following from the Ehrenfeucht–Mostowski theorem) that for every complete theory $T$ and for every $\lambda \geq |T|$, there is $M \models T$ with $|M| = \lambda$ and $M$ having $2^\lambda$-many automorphisms (assuming $T$ has infinite models). So if I'm understanding your question correctly then what you denote as $\mbox{Mod}_\kappa(T)$ always has cardinality $2^{<\kappa}$, at least for $\kappa > |T|$.

PS: (simplified example) To partially address your second question: for every cardinal $\lambda$ (possibly $\lambda$ finite or $0$) there are $M \equiv N$ structures in a countable language such that there are exactly $\lambda$ elementary embeddings from $M$ to $N$. Let $\mathcal{L}$ be the language with infinitely many constant symbols $(c_m: m < \omega)$ and let $T$ say that they are all distinct. For $\lambda > 0$, let $M$ be the model with exactly one unsorted element and let $N$ be the model with exactly $\lambda$ unsorted elements; then there are $n$ elementary embeddings from $M$ to $N$. For $\lambda = 0$ let $M$ have 1 unsorted element and let $N$ have no unsorted elements.

PPS: On the other hand if $M, N$ are countable then the possibilities are exactly $0, 1, 2, \ldots, \aleph_0, 2^{\aleph_0}$. From the above examples we have seen these are all possible, on the other hand the space of elementary embeddings from $M$ to $N$ is a Polish space (completely metrizable space) so either is countable or has a perfect subset. (The subtleties of Vaught's conjecture are that we are looking at models up to isomorphism, so an equivalence relation on a Polish space.)