I think that I can expand on RW's answer

If $I$ is countably infinite, then we can cover $X$ with the countable partition
$$ X = \vee_{n=0}^\infty T^{-n}P $$
Take any element of this partition:
$$Q = P_{n_0} \vee T^{-1}P_{n_1} \vee \cdots \vee T^{-i}P_{n_i} \vee \cdots $$
where $n_i \in [1,k]$. Then
$$
TQ = TP_{n_0} \vee P_{n_1} \vee T^{-1}P_{n_2} \vee \cdots \vee T^{-i+1}P_{n_i} \vee \cdots $$
is contained within partition element
$$ P_{n_1} \vee T^{-1}P_{n_2} \vee \cdots \vee T^{-i+1}P_{n_i} \vee \cdots $$
so $TQ$ (or more generally $T^iQ$) is either disjoint from $Q$ (when it is contained within another partition), or equal to $Q$

There are two possibilities for the measure of $Q$:

(1) the measure of $Q$ is zero, or

(2) the measure of $Q$ is positive, and $T^nQ=T^{n+k}Q$ for some $n,k < \infty$. Otherwise $\{T^iQ\}$ is an infinite sequence of disjoint sets, each of constant measure $\mu(Q)$, and

$$ \infty = \sum_{i=1}^\infty \mu(Q) = \sum_{i=1}^\infty \mu(T^iQ) \leq \mu(X) = 1$$

Hence there are (possibly infinitely many) $Q$'s of finite measure, each with finite orbit.

Each $Q$ represents the set of points that are indistinguishable in your topology, so we can say that the space you are talking about is really just a (countable) union of finite orbits with an atomic probability measure.