TITLE: Probability that a random permutation fixes no elements for large $n$
QUESTION [8 upvotes]: Here's a problem that's been going around my friend group. I have solved this problem, and am interested in a generalization of it.

Here's the original problem: Pick an element of $\sigma\in S_n$ uniformity at random, and let $X(n)$ be the random variable that is the number of fixed points of $\sigma$. Compute 
$$\lim_{n\to\infty}P(X(n)=0)$$

Here's the extension: Suppose instead of $S_n$ we are interested in a class of subgroups, $\{H_n:n\in I\subseteq\mathbb{N}\}$. We can then ask: what is the probability the limit yields
$$\lim_{n\to\infty}\frac{\mu(H_n)}{|H_n|}$$ 
where $\mu(G)$ is the biggest number such that every element of $G$ fixes $\mu(G)$ or more elements of $[n]$ when viewed as a subset of $S_n$. The notion I'm trying to formalize here is "the least number of fixed points that is actually achieved" as for many groups, every element has a fixed point.
This seems to be the right generalization of the problem, though other generalizations are welcome. I'm mostly asking this out of curiosity to see what can be proven. Results about transitive groups, doubly-transitive subgroups, $D_{2n}$, or any other reasonably interesting class of groups would be exciting for me.
This link seems relevant, but it's not clear to me that this problem can be solved simply by summing Recontre numbers. This MSE post also seems relevant, but not quite enough to solve the problem on its own.

REPLY [4 votes]: It is pretty well known that there are
$$ D_n = n! \sum_{k=0}^{n}\frac{(-1)^k}{k!} $$
derangements (permutations without fixed points) in $S_n$, so the limit probability for the first problem is $\color{red}{\large\frac{1}{e}}$. It also is the limit probability for the second case (random even permutation), since the difference between the numbers of even and odd derangements is given by the determinant of a $n\times n$ matrix $M_n$ with zero elements on the diagonal and elements equal to $1$ outside the diagonal. $M_n$ is the difference between a rank-$1$ matrix (having eigenvalues $n,0,0,\ldots$) and the identity matrix, hence the difference between even and odd derangements is $n-1$ in absolute value and the limit probability is the same in both cases.
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