If a linearly ordered set $L$ has the property that every order-preserving injection $L \rightarrow L$ is expansive, is $L$ necessarily well-ordered?

Question

Given a poset $P$, call a function $f : P \rightarrow P$ expansive iff $f(x) \geq x,$ for all $x \in P$. Now suppose a linearly ordered set $L$ has the property that every order-preserving injection $f : L \rightarrow L$ is expansive. Is $L$ necessarily well-ordered?

Answer 1

The answer is no, at least consistently. Under the Continuum Hypothesis, there are uncountable sets of reals $A$ so that any order preserving injection $f:A\to A$ has to be the identity. Such sets are called 2-entangled because of the following equivalent formulation: for any pairwise disjoint family $\mathcal X$ of 2-element subsets of $A$ and any map $s:\{0,1\}\to \{0,1\}$ there is $a,b\in \mathcal X$ so that $a_i<b_i$ iff $s(i)=0$ for $i=0,1$. Here $\{a_0,a_1\}$ is the increasing enumeration of $a$.

2-entangled sets can also exist while Martin's Axiom holds or if one adds $\aleph_1$ Cohen-reals. A good reference might be "On the consistency of some partition theorems for continuous colorings and the structure of $\aleph_1$-dense real order types" by Abraham, Rubin and Shelah.

Edit: here is a ZFC example. In [Lemma 2.3, Baumgartner, James E. "Order types of real numbers and other uncountable orderings." Ordered sets. Springer Netherlands, 1982. 239-277.] a set $Z=\{z_\alpha:\alpha<2^{\aleph_0}\}$ is constructed with the property that if $X,Y\subseteq Z$ are pairwise disjoint of size continuum then $X$ does not embed into $Y$. In short, we list all countable order preserving injections as $\{f_\alpha:\alpha<2^{\aleph_0}\}$ and make sure that $z_\alpha\notin \{\bar f_\xi(z_\zeta),\bar f^{-1}_\xi(z_\zeta):\xi,\zeta<\alpha\}$ where $\bar f_\alpha$ is the largest order preserving map whose values are uniquely determined by $f_\alpha$. It is easy to make sure that $Z$ meets each interval of $\mathbb R$ in $2^{\aleph_0}$ points.

I claim than any order preserving injection $f:Z\to Z$ must be the identity. First, lets prove that $f(x)=x$ for all but $<2^{\aleph_0}$ points. Indeed, otherwise we can find $X,Y\subseteq Z$ pairwise disjoint of size continuum so that $x\in X$ implies $f(x)\in Y$. This contradicts the assumption on $Z$. Now, those points of $Z$ where $f$ is the identity must be dense. So $f$ must be the identity everywhere.