We know that if $f\colon P \longrightarrow Q$ is a bijective poset morphism, then its inverse is not necessarily a poset morphism. If we assume that P and Q are, in addition, linearly ordered, then the statement turns out to be true, that is, $f^{-1}$ is then a poset morphism.
My question is: can we do it better? Do we need to assume that $Q$ is linearly ordered? Is it enough to assume it only for P in order to get that $f^{-1}$ is also a poset morphism?
$f^{-1}$ is a poset morphism iff $f$ is order reflecting, so we need to show that $f(x) \leq f(y) \implies x \leq y$ holds for every $x,y \in P$. Well, take $x,y \in P$. We can assume that $f(x) \neq f(y)$, otherwise by injectivity we are done. So assume that $f(x) < f(y)$ and suppose that $x \nleq y$. Since $P$ is linearly ordered, we must have $y < x$. Using that $f$ is a poset morphism, we get that $f(y) \leq f(x)$. By the antisymmetry property, this could only happen if $f(x) = f(y)$, but this is a contradiction.
Is my reasoning ok? It is curious, because if you try to prove it by contrapositive, it seems that you need also $Q$ to be linearly ordered, so this is kind of showing the difference between these two ways of proving things.
Well, you see, the thing is that if $f:P\to Q$ is bijective and order-preserving, and $P$ is linearly ordered, then so is $Q$.
Indeed, take $u,v \in Q$; then there exist $a,b \in P$ such that $f(a)=u$ and $f(b)=v$ (I think you can tell why!).
Now, since $P$ is linearly ordered, either $a \leq b$ or $b \leq a$; suppose, without loss of generality, that $a \leq b$.
Then, since $f$ preserves order, $u = f(a) \leq f(b) = v$.
(By the way, your reasoning seemed perfectly OK to me.)