Monotone bijections between lists of intervals I have the following problem:

Input: two sets of intervals $S$ and $T$ (all endpoints are integers).
Query: is there a monotone bijection $f:S \to T$?

The bijection is monotone w.r.t. the set inclusion order on $S$ and $T$.
$$\forall X\subseteq Y \in S, \ f(X) \subseteq f(Y)$$

[I am not requiring the reverse condition here. Update: if the reverse condition were required, i.e., $\forall X, Y, X\subseteq Y \Leftrightarrow f(X) \subseteq f(Y)$, then this would be in PTIME because it amounts to isomorphism testing of the corresponding inclusion posets (which have order dimension 2 by construction), which is in PTIME by Möhring, Computationally Tractable Classes of Ordered Sets, Theorem 5.10, p. 61.]

The problem is in $\mathsf{NP}$: we can check efficiently if a given $f$ is a monotone bijection. 


  Is there a polynomial-time algorithm for this problem? 
  Or is it $\mathsf{NP}$-hard? 


The question can be stated more generally as 
existence of a monotone bijection between two given posets of 
order dimension 2. 

Using a reduction inspired by the answers to this question, 
I know that the problem is $\mathsf{NP}$-hard
when dimensions are not restricted.
However, it is not clear 
if the reduction would also work when dimensions are restricted.

I am also interested to know about tractability when 
the dimension is just bounded by some arbitrary constant (not just 2).
Here is an attempt to prove that the problem without the reverse condition is NP-hard.

The basic idea is that disjoint intervals in $S$ like this one:

 [S]  +-a-+ +-b-+
      +---c-----+  c&lt;a, c&lt;b (here &lt; is interval inclusion)


can have a valid mapping to a "pyramid" in $T$:

 [T]  +-x-+      f(a)=x, f(b)=y, f(c)=z
      +-y---+    
      +-z-----+  z&lt;x, z&lt;y OK


The reduction is from Unary 3-Partition (which is NPC).
Given $3m$ integers $A = \{a_1,a_2,...,a_{3m}\}$ and an integer $B$, does exists a partition of A in $m$ sets $A_1,...,A_m$ such that every $A_i$ have exactly 3 elements and their sum is $B$?

Suppose that $max = \sum a_i + 3m$

We construct $S$ adding $3m$ base intervals $BI_i$ of length $3*max$ (red lines in the figure),
on top of each base interval  we add a marker pyramid of $max$ intervals of increasing length (green lines in the figure).
To base interval $BI_i$ we also add $a_i$ disjoint unit intervals of length 1 (black lines in the figure). Finally we add a long interval $L$ to cover all $BI_i$ (blue line in  the figure).

Then we construct $T$ starting from a copy of $L$, then we add $m$ sum groups $G_j$, each one  made with a copy of three stacked base intervals stretched in such a way that their marker pyramids don't intersect (see red+green lines at the bottom of the figure).
Then we add on top of the three base intervals of $G_j$ a sum pyramid of $B$ intervals of increasing length (disjoint from the marker pyramids).

Suppose that there exists a bijection between S and T that preserves the interval inclusion (in one direction from S to T).

Then each marker pyramids of S must correspond to a marker pyramid in T (the only way to have an inclusion chain of $max$ intervals), so exactly three base intervals ($BI_{j_1},BI_{j_2},BI_{j_3}$) of $S$ must be mapped to each group $G_j$. Furthermore, the unit intervals of $BI_{j_k}$ must be mapped to the sum pyramid of $G_j$ and cannot be "exchanged" between different groups. 

In a similar way it can be proved that if there exists a bijection then the original unary 3-partition problem has a solution.


Reduction example from the unary 3-partition problem $m=2, A = \{3,3,2,2,2,2\}, B = 7$

Note: as observed in the comments the blue intervals L in S and T are not essential for the reduction.

If the reverse condition is also required, then you can build two DAGs using the relation $I_i \subseteq I_j$ to build arcs $(I_j \rightarrow I_i)$. A bijection that preserves interval inclusion in both directions exists iif the two DAGs are isomorphic. So the problem cannot be harder than the DAG isomorphism problem, which is GI-complete (and if you prove that it is NP-complete then you prove that GI is NP-complete, too).