What is the probability that two random permutations have the same order? I am interested in the orders of random permutations. Since the law of the logarithm of the order of a permutation converges to a normal law (for instance Erdös-Turan Statistical group theory III), one expects that the probability for two permutations of $\frak S_n$ to have the same order goes to 0 as n goes to infinity. Indeed experimentally this seems to happen with speed $O(1/n^2)$
I know that Wilf proved an asymptotic for a permutation in $\frak S_n$ to be of order $d$ (https://www.math.upenn.edu/~wilf/website/Asymptotics%20of%20exp%28P%28z%29%29.pdf) but I don't think it can be used directly.  
On the other hand it is clear that the probability that two permutations have same order is more than probability that two permutations are conjugate. This is $K/n^2$ according to Flajolet et al. (http://arxiv.org/abs/math/0606370), but here again I failed to generalize the method for the order.
 A: For two random permutations of $n$ letters, let $p_1(n)$ be the probability they are conjugate and $p_2(n)$ be the probability they have the same order.  I computed these exactly up to $n=70$.  In the following, the blue diamonds are $n^2p_1(n)$ and the red circles are $n^2p_2(n)$.
The ratio $p_2(n)/p_1(n)$ is some sort of weighted average of how many different partitions of $n$ have the same lcm, with popular partitions weighted more. I would have guessed this would slowly increase, but that isn't visible. The wriggliness of $p_2(n)$ presumably reflects the fact that the number of partitions with the same order as a given partition is some complicated arithmetic function.
Maple was struggling by the time it got to $n=70$ but a C program should be able to reach $n=100$ or maybe further.

A: Nice problem! I claim that $\limsup n^2 p(n) = \infty$.
Suppose $k < n/2$ is such that $n-k$ is divisible by $L_k = \text{lcm}(1,2,\dots,k)$. Then if $\pi \in S_n$ has a cycle of length $n-k$ (this happens with probability $1/(n-k)$) then $\text{ord}(\pi) = n-k$, so the probability gets a contribution of $1/(n-k)^2 \geq 1/n^2$ from such $\pi$. Let $K_n$ be the set of all such $k$. Then $p(n) \geq |K_n|/n^2$.
Now the great thing about $n \mapsto K_n$ is that it is "lower semicontinuous on $\widehat{\mathbf{Z}}$". What I mean is this: Suppose $K = K_n$, and let $k = \max K$. Then the condition that $K_n \supset K$ is $L_k$-periodic, so provided we alter $n$ only by multiples of $L_k$, the set $K_n$ can only get bigger. Moreover, note that we would have $n \in K_n$, except for our stipulation that $k < n/2$. It follows that
$$
  K_{n + L_n} \supset K_n \cup \{n\}.
$$
This shows that $|K_n|$ gets arbitrarily large, which proves the claim.
I have no idea about the $\liminf$! The above argument constructs very particular $n$ (of the shape $n_1 + \dots + n_k$, where each $n_i$ is large and highly divisible in comparison with $n_{i-1}$), and the lower bound is very weak besides (roughly $\log^*(n)/n^2$), so it's tempting to conjecture that the $\liminf$ is finite. But I'm not sure this is supported by the numerics. From the numerics it just looks like we're forgetting something.
A: That the probability is at least $O(1/n^2)$ is immediate, since the probability that both permutations are $n$-cycles is $1/n^2.$ For the upper bound, there is a convergence speed estimate by Zacharovas, see in particular theorems 3 and 4, which should give you what you want.
