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Assume $A$ is a countable model of $T$ and its automorphism group is an oligomorphic permutation group. Complete types are constant on automorphism orbits, so only finitely many complete $n$-types are realized in $A$, for each $n$.

These realized types are dense in the Stone topology on a type space $S_n(T)$. Indeed, any nonempty basic clopen set $[\theta ]$ is given by a formula consistent with $T$. Completeness implies $T\models \exists \mathbf{x}\theta$, and thus some tuple in $A$ realizes a type in $[\theta ]$. A finite set of points is closed in a Hausdorff space; if it is also dense, it is the whole space. Therefore $S_n(T)$ equals the finite set of types realized in $A$. This proves (iii) implies (ii), and the previous part then yields (i). We have establishedIn particular, the argument does not assume that an arbitrary countable model is saturated or that its type-orbits coincide before that coincidence has been proved.

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).