I have a proof of the following:
[*] Let $A$ be a countable $\omega$-saturated model of a complete, countable theory $T$ (with infinite models). There is a bijection between the orbits under the action of $\text{Aut}(A)$ on $A^n$ and the types $S_n(T)$.
Here the orbit of $(a_1, \dots, a_n)$ under the action of $\text{Aut}(A)$ on $A^n$ is defined as the set $\{(f(a_1), \dots, f(a_n)) : f \in \text{Aut}(A)\}$.
I want to use this to prove:
If for every countable model $A \models T$ and $n \in \mathbb{N}$, the set of orbits under the action of $\text{Aut}(A)$ on $A^n$ is finite, then $T$ is $\omega$-categorical. (The reverse is not so difficult.)
My attempt so far: it suffices to find a countable $\omega$-saturated model of $T$, because then the result follows from the assumptions, lemma [*], and the theorem $\omega$-categorical $\Leftrightarrow$ every $S_n(T)$ is finite. Such a model exists iff every $S_n(T)$ is countable, so it would be enough to prove that $T$ has only countably many countable models. I thought this would be easy enough, but then I found out that there are theories with a countable $\omega$-saturated model that have continuum-many countable models. So things are a bit more complicated...
If you already know that $\omega$-categoricity is equivalent to the condition that every $S_n(T)$ is finite, then there's an easy argument that doesn't use [*].
Suppose for contradiction that there is some $n\in \omega$ such that $S_n(T)$ is infinite. Pick some countable set of these and realize them: Then you have a countable model $A\models T$ which realizes infinitely many $n$-types. But $n$-tuples which realize different types cannot be in the same orbit of the action of $\text{Aut}(A)$ on $A^n$, so there are infinitely many orbits, contradiction.
The point is that [*] is really two statements:
(1) If $a$ and $b$ are in the same orbit, they realize the same type.
(2) If $a$ and $b$ realize the same type, they are in the same orbit.
Point (1) is true in any model, while point (2) is only true in certain models, e.g. saturated ones. And the argument above only uses (1), not (2).