A sufficient condition for order isomorphism of posets? Let $\mathfrak{A}$ be a poset. For $a, b \in \mathfrak{A}$ we will denote $a
\not\asymp b$ if only if there are a non-least element $c$ such that $c
\leqslant a \wedge c \leqslant b$.
Let $\mathfrak{A}$, $\mathfrak{B}$ are posets. I call a pointfree
funcoid a pair $\left( \alpha ; \beta \right)$ of functions $\alpha :
\mathfrak{A} \rightarrow \mathfrak{B}$, $\beta : \mathfrak{B} \rightarrow
\mathfrak{A}$ such that
$$ \forall x \in \mathfrak{A}, y \in \mathfrak{B}: \left( y
   \not\asymp^{\mathfrak{B}} \alpha \left( x \right) \Leftrightarrow x
   \not\asymp^{\mathfrak{A}} \beta \left( y \right) \right) . $$
Conjecture
  If $\left( \alpha ; \beta \right)$ is a pointfree funcoid and $\alpha$ is a
  bijection $\mathfrak{A} \rightarrow \mathfrak{B}$, then $\alpha$ is an order
  isomorphism $\mathfrak{A} \rightarrow \mathfrak{B}$.
A weaker conjecture:
Conjecture
  If $\left( \alpha ; \beta \right)$ is a pointfree funcoid and $\alpha$ is a
  bijection $\mathfrak{A} \rightarrow \mathfrak{B}$ and $\beta$ is a bijection
  $\mathfrak{B} \rightarrow \mathfrak{A}$, then $\alpha$ is an order
  isomorphism $\mathfrak{A} \rightarrow \mathfrak{B}$.
If these conjectures are false, what additional conditions we may add to make them true? (Maybe, these are true for lattices? distributive lattices?)
 A: Both conjectures are false for infinite posets. Take for instance any bijections of sets $\alpha,\beta$ ($\beta=\alpha^{-1}$ works fine) between $\mathbb{Z}$ and $\mathbb{Q}$. Because for all $m,n\in\mathbb{Z}$ and for all $r,s\in\mathbb{Q}$ we always have $m\asymp n$ and $r\asymp s$, there is really no condition imposed on $\alpha$ and $\beta$, but neither can be an order isomorphism.
Maybe the answer is affirmative if you look at finite posets.
A: Let $\mathfrak{A}$ is a poset, let $\star a=\{ x\in\mathfrak{A} | x\not\asymp a \}$.
I will call a poset $\mathfrak{A}$ separable when $\star a=\star b \Leftrightarrow a=b$ for every $a,b\in\mathfrak{A}$.
Theorem
  Let $\left( \alpha ; \beta \right)$ is a pointfree funcoid from a separable poset
  $\mathfrak{A}$ to a separable poset $\mathfrak{B}$. If $\alpha$ is an injection, then
  $\alpha$ is an order embedding $\mathfrak{A} \rightarrow \mathfrak{B}$.
Proof Suppose $x \geqslant y$ but $\alpha (x) \ngeqslant
\alpha (y)$.
Then by separability of $\mathfrak{B}$ there exist $z \not\asymp \alpha (y)$ such
that $z \asymp \alpha (x)$.
Thus $\beta (z) \asymp x$ and $\beta (z) \not\asymp y$ what is impossible for $x
\geqslant y$. $\Box$
Corollary
  Let $\left( \alpha ; \beta \right)$ is a pointfree funcoid from a separable poset
  $\mathfrak{A}$ to a separable poset $\mathfrak{B}$. If $\alpha$ is a bijection
  $\mathfrak{A} \rightarrow \mathfrak{B}$, then $\alpha$ is an order
  isomorphism $\mathfrak{A} \rightarrow \mathfrak{B}$.
