Suppose $S$ is a finite semigroup. Let's define the commuting fraction $cf(S) := \frac{|\{(a, b) \in S^2| ab = ba\}|}{|S|^2}$. Alternatively, it can be defined as the probability that two independent uniformly distributed random elements of $S$ commute.
Now, suppose, $A$ is a finite set, $|A| = n$, $F_n$ is the semigroup of all functions $A \to A$ under composition. What is the asymptotics of $cf(F_n)$?
What have I attempted so far:
Let's for $g: A \to A$ define $St(g) := \{a \in A| g(a)=a\}$. Then it is not hard to see, that if $St(a) \cup St(b) = A$, then $a$ and $b$ commute. Thus $cf(F_n)$ is greater than probability that $St(a) \cup St(b) = A$. For any given set $B \subset A$, the probability, that $St(a)=B$ and $A \setminus B \subset St(b)$ is equal to $(\frac{1}{n})^{|B|}(\frac{n-1}{n})^{n-|B|}(\frac{1}{n})^{n-|B|} = (\frac{1}{n})^{n}(1 - \frac{1}{n})^{n-|B|}$. That means, that the probability of $St(a) \cup St(b) = A$ is $\sum_{k = 0}^n C_n^k (\frac{1}{n})^n(1 - \frac{1}{n})^{k} = (\frac{1}{n})^n(2 - \frac{1}{n})^n \geq e^{-n\ln(n)}$. Thus $cf(F_n) = \Omega(e^{-n\log(n)})$.
This bound seems to be quite crude and one-sided. However I found neither any better lower bound nor any non-trivial upper bound.