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Assume every $S_n(T)$ is finite. Each of its points is isolated: choose, for every other type, a formula separating it from the given one, and conjoin the finitely many choices. Thus every finite tuple in every model has an isolated complete type. Moreover every complete type is realized in every model: its isolating formula is consistent with the complete theory, so the theory entails that the formula has a realization.

For clarity, here is the one-point extension argument needed for both remaining assertions. If tuples $\mathbf{a}$ and $\mathbf{b}$ in two models have the same complete type and $c$ is any new element in the first, let $\theta (\mathbf{x},y)$ isolate $\mathrm{tp}(\mathbf{a},c)$. The formula $\exists y\theta (\mathbf{x},y)$ belongs to the type of $\mathbf{a}$, hence also to that of $\mathbf{b}$. Choose $d$ with $\theta (\mathbf{b},d)$. This extends the tuple map to a partial elementary map. If $c$ is different from all earlier entries, the isolating type includes these disequalities, so $d$ is new as well.

Apply this argument alternately to the two enumerations of any two countable models. The resulting back-and-forth map is an isomorphism. Thus (ii) implies (i).

Choose any countable model $A$. The same back-and-forth argument, now within $A$, shows that any two tuples of the same complete type are related by an automorphism of a first-order structure. Conversely an automorphism preserves the type. Hence the orbits of $\mathrm{Aut}(A)$ on $A^n$ are exactly its complete $n$-types. There are finitely many of them for every $n$, so the automorphism group is an oligomorphic permutation group. Therefore (ii) also implies (iii).

Expand the language by constants for $A$ and write $T_A=\mathrm{Th}(\mathcal{M},a{)}_{a\in A}$. An A-type over a structure is a set of formulas in a fixed finite tuple $\bar{x}$, with parameters from $A$, consistent with this complete theory.

Let $Q=\{{\phi }_1(\bar{x}),\dots ,{\phi }_m(\bar{x})\}\subseteq P$. If no tuple of $M$ realized $Q$, thenThat sentence belongs to $T_A$ and contradicts the consistency of $T_A\cup P$. ConsequentlyFor $Q=\varnothing$, any tuple works because domains are nonempty. This proves finite satisfiability; realization of the entire partial type may require an elementary extension.

Let $p(\bar{x})$ be an isolated type over $A$, with isolating formula $\theta (\bar{x})$ over $A$. Its consistency with $T_A$ and the complete theory property implyotherwise $\neg \exists \bar{x}\theta$ would belong to $T_A$, contradicting consistency. Choose a tuple $\bar{b}\in M$ satisfying $\theta$. By isolation,so $\bar{b}$ satisfies every member of $p$. Therefore every isolated A-type is realized in the original structure. The same reasoning works for a principal partial type whose isolating formula need not itself be a member of the type.