locally incomparable dense linear orderings extending $\langle \mathbb{R}, < \rangle$ This follows up on Incomparable dense linear orderings extending $\langle \mathbb{R},< \rangle$
and hopefully sparks more discussion.
Where $a<b$, say that the four “types” of nonempty bounded intervals are:
$(a,b)$, $[a,b]$, $(a,b]$, and $[a,b)$.
Let $\langle X,< \rangle$ and $\langle Y,< \rangle$ be dense linear orderings without endpoints such that 


*

*$\langle \mathbb{R},< \rangle$ is order-embeddable into $\langle X,< \rangle$, and into $\langle Y,< \rangle$,

*all the nonempty bounded intervals of $\langle X,< \rangle$ of the same type are order-isomorphic,

*all the nonempty bounded intervals of $\langle Y,< \rangle$ of the same type are order-isomorphic,

*$\langle X,< \rangle$ is isomorphic to its inverse order, and

*$\langle Y,< \rangle$ is isomorphic to its inverse order.
Are there examples of $\langle X,< \rangle$ and $\langle Y,< \rangle$ such that

“no nonempty open interval of $\langle X,< \rangle$ is order-embeddable into any nonempty open interval of $\langle Y,< \rangle$, and no nonempty open interval of $\langle Y,< \rangle$ is order-embeddable into any nonempty open interval of $\langle X,< \rangle$”?

Is it independent of $ZFC$ whether there are such pairs of orderings for which the statement in quotes holds?
If $ZFC$ proves that there are such pairs of orderings, then which definitions (if any) of such orderings $\langle X,< \rangle$ and $\langle Y,< \rangle$ are such that it is independent of $ZFC$ whether the statement in quotes holds of them?
 A: Here's a slight variation on Hamkins' example using compactness theorem under the assumption $2^{\kappa} > \kappa^+$ for some $\kappa \geq \mathfrak{c}$. Let $T$ be the (consistent) theory consisting the elementary diagram of $(\mathbb{R}, <)$ and the following sentences:
(1) $c_i < c_j$ for each $i < j < \kappa^{+}$
(2) For all $a < b$, the map $x \mapsto f(a, b, x)$ is an order preserving bijection from $(a, b)$ to the universe
(3) $x \mapsto g(x)$ is a reverse order isomorphism of the universe
By Lowenheim-Skolem, $T$ has a model $X$ of size $\kappa^{+}$. Let $Y = {}^{\kappa}\mathbb{R}$ under lexicographic order. Then neither one of $X, Y$ embeds in the other - $X$ has an increasing $\kappa^+$-sequence while $Y$ has size $|Y| = 2^{\kappa} > \kappa^+ = |X|$.
A: Let me prove at least that it is consistent with ZFC that there are
two linear orders as you requested.
Work in ZFC plus the assumption that
$2^{\omega_1}>\omega_3^L$, but $\omega_2=\omega_2^L$ and
$\mathbb{R}=\mathbb{R}^L$. This situation is easy to arrange by
forcing, since you can simply add $\omega_4$ many Cohen subsets of
$\omega_1$ over $L$, which preserves cardinals and adds no reals.
Under that assumption, here are two linear orders with the
properties you have requested. Let $X=(\mathbb{Q}^{\omega_2})^L$
and $Y=\mathbb{R}^{\omega_1}$, each under the lexical order.
The real line embeds into both of these orders. Both of these
orders have the all-intervals-look-alike property. And both of the
orders is symmetric, in the sense of being isomorphic to its own
inverse order.
Meanwhile, every nontrivial interval of $X$ has an increasing
$\omega_2$-sequence, and there is no such sequence in $Y$, so no
interval of $X$ embeds into $Y$.
Conversely, every interval of $Y$ has size $2^{\omega_1}$, which is
larger than $X$, since $X$ has size merely $(2^{\omega_2})^L=\omega_3^L$, and so
no interval of $Y$ embeds into $X$.
