Here are necessary and sufficient conditions for an abstract, finite simplicial complex $\mathcal{S}$ to be the order complex of some partially ordered set.

(i) $\mathcal{S}$ has no missing faces of cardinality $\geq 3$; and

(ii) The graph given by the edges (=$1$-dimensional simplices) of $\mathcal{S}$ is a comparability.

[Definitions.
(a) A *missing face* of $\mathcal{S}$ is a subset $M$ of its vertices (=$0$-dimensional simplices) such that
$M \not \in \mathcal{S}$, but all proper subsets $P\subseteq M$ satisfy $P\in \mathcal{S}$.
(b) A graph (=undirected graph with no loops nor multiple edges) is a *comparability* if its
edges can be transitively oriented, meaning that whenever edges $\{p, r_1\}, \{r_1, r_2\},\ldots, \{r_{u−1}, r_u\}, \{r_u, q\}$ are oriented as $(p, r_1), (r_1, r_2),\ldots, (r_{u−1}, r_u), (r_u, q)$, then there exists an edge $\{p, q\}$ oriented as $(p, q)$.]

This characterisation appears with a sketch of proof $-$ which is not hard, anyway $-$ in

M. M. Bayer, *Barycentric subdivisions*. Pacific J. Math. 135 (1988), no. 1, pp. 1-16.

As Bayer points out, the result was first observed in

R. Stanley, *Balanced Cohen-Macaulay complexes*, Trans. Amer. Math. Soc, 249 (1979), pp. 139-157.

@Rasmus and @Gwyn: The characterisation might perhaps disappoint you if you were expecting something more topological. However, it's easy to prove that no topological characterisation of order complexes is possible, and therefore a combinatorial condition such as the one on comparabilities *must* be used. For this, first check that the barycentric subdivision of *any* simplicial complex indeed is an order complex. Next observe that barycentric subdivision of a simplicial complex does not change the homeomorphism type of the underlying polyhedron of the complex. Finally, conclude that for any topological space $T$ that is homeomorphic to a compact polyhedron, there is an order complex whose underlying polyhedron is homeomorphic to $T$.

I hope this helps.

baryzentric subdivision ofthe triangulation we started with. $\endgroup$