Isomorphic dense linear orders? 
Question: Suppose $(A,<)$ and $(B,<)$ are two dense linear orders without end points, of size $2^{\aleph_0}$, and both $\aleph_1$-saturated (every
  type over a countable subset has a realization). Are they isomorphic?

I know that under the continuum hypothesis, one can use a back & forth argument because it is true that any two $\kappa$-saturated models of size $\kappa$ have to be isomorphic. However, I don't know what happen in the general case. 
Are there examples assuming $\neg CH$?
 A: Suppose we have $2^{\aleph_0}>\aleph_1$.
Hausdorff showed that there is an $\eta_1$-order without end points of size $2^{\aleph_0}$, that is, a totally ordered set $\Bbb A=(A,<)$ having the following properties:


*

*$\Bbb A$ has neither a coinitial nor a cofinal subset of size $<\aleph_1;$

*For any $B,C$ subsets of $A$ both of size less than $\aleph_1$ with $B<C$, there is some $a\in A$ with $B<a<C$.


$\eta_1$-orders without end points are $\aleph_1$-saturated models; see section $5.4$ of Chang & Keisler's Model Theory, third edition.
Let $<_1$ be the lexicographical order on $\omega_1\times A$, and set $\Bbb A_1=(\omega_1\times A,<_1)$.
Build another total order $\Bbb A_2$ just like $\Bbb A_1$, but instead of using $\omega_1$, use $\omega_2$. 
As $A$ has no end points, it is easy to see both $\Bbb A_1$ and $\Bbb A_2$ are $\eta_1$-orders. These orders have no end points, thus they are $\aleph_1$-saturated, and have size $2^{\aleph_0}$. However, they cannot be isomorphic as $\Bbb A_1$ has a cofinal subset of size $\aleph_1$, while $\Bbb A_2$ does not.
