Is there any finite self-dual lattice $(X,\le)$ such that there is not any self-duality $f:X\to X$ such that $f\circ f = 1_X$?
Let $f,g:X\to X$ be a self-dualities. Then $f^{-1}\circ g$ is an order-isomorphism. So $g=f\circ (f^{-1}\circ g)$. This means that the set of all self-dualities is of the form $f\circ \theta$ where $\theta$ is an order-isomorphism. The question s if there's any order-isomorphsim $\theta$ such that $$f\circ \theta\circ f\circ \theta = 1$$
There is a smallest $n$ with $$f^{2n}=1_X$$ If it can be proved that $n$ is odd then $f^n$ is a self-duality and $(f^n)^{2}=1_X$
There do exist finite self-dual lattices with no self-duality of order $2$. I will draw a picture of one of size $34$ whose self-dualities all have order $8$.
The lattice is
On the left you find the Hasse diagram of the lattice with some directed edges. The upward direction on the edge $a-A$ is meant to indicate that the edge really represents an interval isomorphic to the $5$-element non-self-dual lattice $Z$ which appears on the right. I use the asymmetry of $Z$ to encode orientations into some edges. Each directed edge in the left lattice is meant to represent a copy of $Z$ in one orientation or the other, depending on the direction of the arrow.
Any self-duality of the lattice on the left must interchange atoms and coatoms and preserve the orientations of the oriented edges. Therefore $A$ must be mapped to one of $a, b, c$ or $d$, and once that is decided the entire map is determined. The self-dualities turn out to be exactly the odd powers of the $8$-cycle $(A\;a\;B\;b\;C\;c\;D\;d)$. The order of any of these is $8$.