uniqueness of order-isomorphisms for a finite lattice

Question

Let $(X,\le)$ be a finite lattice and $f,g:(X,\le)\to (X,\le^{-1})$ be order-isomorphisms. Is $f$ the same as $g$‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌?

Answer 1

The answer is no.

A counter example would be the two point discrete-ordered set: {a,b} with ordering $\leq \,:=\, \{(a,a) , (b,b)\}$. Then the identity $x \mapsto x$ and swapping $a \mapsto b, b \mapsto a$ are order-isomorphisms between the ordered-set and the converse-ordered set. However they are distinct isos.

This was pointed out to me by a friend; after which he exclaimed:

``Totally ordered sets are boring! They don't exist in real-life!''

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EDIT :: I forgot that we needed a lattice. Fine: consider the diamond: ${\bot,a,b,\top}$ where $\bot$ is bottom element, $\top$ is top element, and $a,b$ are incomparable. ( I do not know how to draw a diamond here ... )

Now the meet of any two non-$\top$ elements is $\bot$; and $\top$ is unit of meet.

Also the join of any two non-$\bot$ elements is $\top$; and $\bot$ is unit of join.

Hence, we've a lattice.

Now consider the distinct maps $$f : \top \mapsto \bot, a \mapsto a, b \mapsto b, \bot \mapsto \top$$ and $$g : \top \mapsto \bot, a \mapsto b, b \mapsto a, \bot \mapsto \top$$

Indeed these are appropriate order-isos :)

Answer 2

The lattice automorphisms of $(X,\le)$ and $(X,\le^{-1})$ are the same functions $X\to X$.

If $f:(X,\le)\to (X,\le^{-1})$ is any order isomorphism and $\alpha$ a nontrivial automorphism of either, then $f\circ\alpha$ and $\alpha\circ f$ are also isomorphisms and are distinct from $f$. So your problem comes down to determining if lattices can have nontrivial symmetries. The answer is yes - the simplest instance is the diamond provided in Moses' answer: the symmetry simply swaps the two incomparable elements whilst fixing the top and bottom ones.