Let $(X,\leq)$ be a pospace: a topological poset such that uppersets and downsets are closed in the topology. Let $A\subseteq X$ be a closed (connected) subspace and suppose $\phi: A\rightarrow A$ is an order isomorphism: a monotone bijection whose inverse is also monotone. Does $\phi$ extend to an order isomorphism of $X$? If not, what kind of conditions on $X$ do guarantee this property?
I'm specifically interested in the case where $X=[0,1]^n$ in which case things like Tietze's extension theorem might be helpful (or just the general fact that it is a compact metric space).
I found a counter-example to the case of $X=[0,1]^n$:
Pick the subspace $A=\frac{1}{2}+[0,\frac{1}{2}]^n$. So it is an upper corner that is order-isomorphic to $X$. One of the corners of $A$ is an interior point of $X$, while the other corners of $A$ are also corners of $X$. If we permute the coordinates of $A$ such that one of the corner points of $A$ is mapped to the "interior" corner point, then we get an order isomorphism of $A$ that can't be extended to an order isomorphism of $X$, since any order isomorphism of $X$ must map corner points to corner points.
Since this applies to a connected compact subset of a compact metric space, it is unlikely to hold in general for any other general class of pospaces.