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...