There are $n$ men, standing one at each vertex of a convex $n$-gon. If they are allowed to move together along sides or diagonals of the polygon to reach another vertex, how many different ways are there to do so without meeting another one?
See OEIS A350599 for the first few numerical values.
Assume that the paths may not cross and each man must move. Label the vertices $1,2,\dots,n$ in clockwise order. Let the man at vertex $i$ move to vertex $\pi(i)$, so $\pi$ is a permutation of $1,2,\dots,n$. If we draw an arrow from vertex $i$ to $\pi(i)$, then we get a disjoint union of noncrossing cycles of length $\geq 3$. We can obtain such cycles by choosing a noncrossing partition of the vertices with no blocks of size 1 and 2, and then orienting the boundary of the convex hull of each block in two ways. Thus in Exercise 5.35(b) of Enumerative Combinatorics, vol. 2, we should set $f(i)=2$ for $i\geq 3$ and $f(1)=f(2)=0$. If the desired answer is $h(n)$, then by this exercise we have $$ x+\sum_{n\geq 1}h(n)x^{n+1} = \left( \frac{x}{1+2\sum_{n\geq 3} x^n}\right)^{\langle -1\rangle} $$ $$ = \left( \frac{x(1-x)}{(1+x)(1-2x+2x^2)}\right)^{\langle -1\rangle}, $$ where $\langle -1\rangle$ denotes compositional inverse.
If a man is allowed to stand still, replace $1+2\sum_{n\geq 3}x^n$ by $1+x+2\sum_{n\geq 3}x^n$.
Possibly you can get some kind of explicit formula for $f(n)$ out of this, but it will be messy.
