An exercise in Stanley's Enumerative Combinatorics (Chapter 3, ex. 8) asked for an example of a finite self-dual poset, (i.e. there is a bijection $f: P\to P$ such that $s\le t \Longleftrightarrow f(s) \ge f(t)$) for which there is no such bijection $f$ satisfying $f(f(t)) = t$ for all $t \in P$.
I'm interested in how many of these such posets exist for a given order. Has there been any result obtained about this? It seems like an interesting problem, but I struggled to find a single one (I would put the Hasse diagram here, but I'm not good enough at $\LaTeX$ to do that) , much less count them.
I posted this question on Math.SE, but it didn't get much of a response, so perhaps the question is better suited for this website.