= Solution
Since the <hom-sets> in these <categories> are either empty or singleton, an <adjunction> $f\dashv g$ is precisely the condition
$$
\boxed{f(x)\le y\quad\Longleftrightarrow\quad x\le g(y)\qquad(x\in X,\ y\in Y).}
$$
The <hom-set> <bijections> are then the unique possible ones, and <naturality> is automatic. This is a <Galois connection> between the two <posets>.
Equivalently, the monotone <functions> satisfy the unit and counit inequalities $x\le gf(x)$ and $fg(y)\le y$. The equivalence implies these by taking $y=f(x)$ and $x=g(y)$. Conversely, if the inequalities hold, $f(x)\le y$ implies $x\le gf(x)\le g(y)$, while $x\le g(y)$ implies $f(x)\le fg(y)\le y$.
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