For $w$ a permutation, let $c(w)$ denote the total number of cycles and $c_i(w)$ denote the number of $i$-cycles. The Ewens measure is a one-parameter probability distribution on permutations where the permutation $w \in S_n$ has probability $$ \frac{\theta^{c(w)}}{n! \cdot \binom{n+\theta-1}{\theta-1}} $$ of occurring. Here, the weight of $w$ is $\theta^{c(w)}$ while the denominator is the normalizing constant. Let $X^\theta_n$ denote a permutation in $S_n$ sampled from the Ewens measure with parameter $\theta$.
A result of Watterson shows that as $n \to \infty$, we have $$c_i(X^\theta_n) \xrightarrow{d} Poi(\theta/i),$$ where $\xrightarrow{d}$ denotes convergence in distribution. A proof with specific rates of convergence appears here.
Note for fixed $n$ if $\theta \to 0$ that the Ewens measure converges to the uniform measure on cyclic permutations, while if $\theta \to \infty$ it converges to the atomic measure on the identity permutation (all fixed points). I am interested in the case where $\theta$ is treated as a function of $n$, denoted $\theta(n)$. When $\theta(n) \to 0$ or $\theta(n) \to \infty$, the limiting behavior now depends on rates of convergence.
Question: Has anyone studied cycle counts for the Ewens measure when $\theta$ is a function of $n$?
I am especially interested in understanding the (normalized) distribution when $\theta(n) \to \infty$. For example, in this case, when is $$ \mathbb{E}\left[c_1(X^{\theta(n)}_n)\right] = o(n)? $$ Naively, the Poisson convergence might suggest this holds for $\theta(n) = o(n)$, but I don't know an easy way to see what should actually be true.