Any theory about period length for AES applied to itself?
Since AES under any fixed key is a permutation, we necessarily have $j = 0$ and $i = l$—iterating a permutation enough times will always return you to the starting point.

From Harris 1960 (paywall-free), if we model AES as a uniform random permutation, every period length $l$ has equal probability $1/n$ (Eq. 5.2) for any particular starting point, where $n = 2^{128}$ is the size of the domain, so the expected cycle length is $\sum_{i=1}^n i/n = (n + 1)/2 \approx 2^{127}$.

(Any substantial deviation from this would imply an attack on AES.)