A Suslin line is a linear order $L$ which is dense with no endpoints, complete, and ccc but not separable. I'm wondering what kind of order-preserving maps there are from $L$ into $L$. Specifically,

**Question**: Can there be a Suslin line $L$ such that for every one-to-one, monotonic function $f$ from an uncountable subset of $L$ into $L$, the set of $x\in\mathrm{dom}(f)$ with $f(x)\neq x$ is countable?

The motivation for this question is the following. A linear order $L$ is *$n$-entangled* if for every uncountable set $A$ of pairwise-disjoint $n$-tuples in $L$, and for every $s : n\to 2$, there are $a,b\in A$ such that $a_i < b_i$ if and only if $s(i) = 0$, for all $i < n$. One can show that a linear order $L$ is $2$-entangled if and only if it is rigid, in the sense that every one-to-one, monotonic function on an uncountable subset of $L$ is equal to the identity on a co-countable subset of its domain.

It's not difficult to show that a weakening of the Open Coloring Axiom implies there are no $2$-entangled sets of reals. However, OCA is consistent with the existence of a Suslin line. So an answer to the above question would provide evidence for an answer to the following.

**Question**: Is OCA consistent with the existence of a $2$-entangled linear order?

**Edit**: I forgot to mention why entangledness is relevant. A $3$-entangled linear order is necessarily separable, so $2$-entangledness is the most you might possibly get out of a Suslin line.