Let $S_n$ be the set of permutations of the first $n$ integers.
Draw at random an element $\pi \in S_n$ with probability
$$
P(\pi) = \frac{1}{Z_{N,n}}N^{\mathcal{L}(\pi)},
$$
where $\mathcal{L}(\pi)$ is the number of cycles of the permutation $\pi$ and $N >0$ is a parameter, $Z_{N,n}$ is a normalisation constant.
Define the match set as the set of integers belonging to a cycle of length one, $M(\pi) = \{i \in \{1,2 … n \} : \pi(i)=i \}$( i.e, the fixed point of the permutation).
How does the expected size of the match set,
$$
E_{N,n}(| M| )
$$
qualitatively grow with n for any value of $N \geq 1$? Here $E_{N,n}$ is the expectation with respect to the probability measure above.
I am particularly interested in the case of integers $N \geq 2$.
The problem is very interesting on its own, for N=1 the problem is well known and $E_{N,n}(| M| ) = 1$ for any $n \in \mathbb{N}$, see for example https://golem.ph.utexas.edu/category/2019/12/random_permutations_part_6.html.
Best Answer
The normalization constant is in fact $Z_{N,n} = N(N+1) \cdots (N+n-1)$. Let $X \subset S_n$ be the set of permutations fixing some point, say $n$. Now elements of $X$ are like elements of $S_{n-1}$ except they have one more cycle, so the total Ewens mass in $X$ is $N Z_{N,n-1} = N\cdot N(N+1)\cdots(N+n-2)$, so $P(X) = N/(N+n-1)$. By linearity of expectation the expected number of fixed points is $nN/(N+n-1)$.
Note this has the right behavior at $N=1$ (uniform distribution), $N = 0$ (uniform distribution on $n$-cycles), and $N = \infty$ (trivial distribution), so "it must be right".
This is fairly standard and there are similar estimates for cycles, fixed sets, etc, all generalizing properties of the uniform distribution. Have a look at Section 3 of this paper if you are interested: https://arxiv.org/abs/1808.08892