The theory of Dense Linear Orders Without Endpoints (DLOWE) is countably categorical, but this is not true for other cardinalities. Let's look at $\beth_1$.
There are at least two order types our set could have.
- $(\mathbb{R}, <)$.
- $(\mathbb{R} \setminus \{0\}, <)$ (as pointed out by Noah Schweber in this comment)
And it's clear that there are many more, such as $\mathbb{R} \setminus \mathbb{Z}$.
Is there a classification of all of them?
(As a quick note, I completely changed this question (which is, I think, permissible since it didn't have any answers))
The $\aleph_0$-categoricity of DLOWE is misleading. In fact, DLOWE is unstable and so in general has as many models of a given uncountable cardinality $\kappa$ as it possibly could, namely $2^\kappa$. This (and vastly more) was proved by Shelah; the key term is "classification theory."
To see how this can be the case, here's an analysis for $\kappa=\aleph_1$. We want to assign to each $S\subseteq\aleph_1$ a DLOWE $L_S$ such that $S\not=T\implies L_S\not\cong L_T$. For simplicity I'll assume $\mathsf{CH}$ here, although it's not hard to remove. The following construction then works. First, we set $$\mathbb{R}_{\alpha,S}=\begin{cases} \mathbb{R} & \mbox{ if }\alpha\in S,\\ \mathbb{R}\setminus\{0\} & \mbox{ if }\alpha\not\in S,\\ \end{cases}$$ and let $$L_S=\sum_{\alpha\in\omega_1}(\mathbb{Q}+\mathbb{R}_{\alpha,S}+\mathbb{Q}).$$
If you want to get rid of $\mathsf{CH}$ here, you just need to be a bit more careful about the coding process used. One way to do this is to let $A$ be an $\aleph_1$-dense subset of $(0,1)$ of cardinality $\aleph_1$, and use as our "coding blocks" the linear orders $1+A$ and $A+1$, the point being that we can tell which side the "$1$" is on. And this basic strategy will work for any uncountable cardinal in $\mathsf{ZFC}$ alone.