Is there any self-dual lattice $(X,\le)$ such that there is **not** any self-duality $f:X\to X$ such that $f\circ f = 1\_X$?



Self-duality in a lattice
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.


