Is there any selfdual lattice $(X,\le)$ such that there is not any selfduality $f:X\to X$ such that $f\circ f = 1_X$?
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Yes. Let $L$ be the lattice structure on $\mathbb Z$ with the following Hasse diagram:
6 < 2 < 2 < 6 <
\ / \ / \ / \
... 5 3 1 1 3 5 7 ...
\ / \ / \ / \
> 4 > 0 > 4 > 8
where all the diagonal arrows go upwards. It is easy to see that the only selfdualities of $L$ are of the form $f(n)=n+c$ for $c\equiv2\pmod4$, and in particular, they are never involutive.

$\begingroup$ Is there any finite counterexamples? Or an infinite complete lattice? $\endgroup$ – Minimus Heximus Aug 4 '14 at 13:34

$\begingroup$ I don’t know. A finite selfdual lattice must have a selfduality whose order is a power of $2$, but I see no particular reason there should be an involution. On the other hand, I don’t see how to construct a counterexample. $\endgroup$ – Emil Jeřábek Aug 4 '14 at 14:16

1$\begingroup$ In view of your other question, my example can be modified by adding a top element, a bottom element, and a “middle” element separating the two $\mathbb Z$chains. Then it becomes an algebraic complete lattice, hence (by a result of Tůma) isomorphic to an interval on the subgroup lattice of some group. $\endgroup$ – Emil Jeřábek Aug 4 '14 at 14:51

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