Self-duality in a lattice 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 $(f^n)^{2}=1_X$
 A: 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 $8$.
