Self-embeddings of uncountable total orders, 2 Let $S = (\Omega,\leq)$ be an uncountable dense total order, such that for all positive integers $m$ and all finite ordered sequences $a_1 < a_2 < \ldots < a_m$ and $b_1 < b_2 < \ldots < b_m$, we have an order automorphism of $S$ mapping the former sequence to the latter. 
Does there exist a proper subset $\Omega' \subset \Omega$ such that the induced total order $S'$ is order isomorphic to $S$ ?
If so, can $\Omega'$ be chosen to be an interval ? 
(I guess the answer will follow from some general theorem on total orders, and I would love to know what that is.)
Thanks !
 A: For the first question, the answer is YES, even if we just assume that $S$ is not rigid. Indeed, fix $f\in\mathrm{Aut}(S)$ and $a\in S$ such that $a<f(a)$. Then $\bigl(\mathrm{id}\restriction(-\infty,a]\bigr)\cup\bigl(f\restriction(a,+\infty)\bigr)$ is an isomorphism of $S$ to its proper suborder $(-\infty,a]\cup(f(a),+\infty)$.
For the second question, the answer is NO in general. One counterexample is the lexicographic product $S=(\omega_1^*+\omega_1)\times\mathbb Q$.
To see that $S$ has the required property (which, by the way, is what model theorists call being strongly $\omega$-homogeneous), notice that any given $a_1,\dots,a_m,b_1,\dots,b_m$ are included in a countable open subinterval of $S$. Such a subinterval must be isomorphic to $\mathbb Q$, which is strongly $\omega$-homogeneous, and an automorphism of the subinterval extends to an automorphism of $S$ by extending it with the identity.
On the other hand, if $S'\subseteq S$ is isomorphic to $S$, it has upwards and downwards cofinality $\omega_1$, which can only happen if it is an (upwards and downwards) cofinal subset of $S$. Thus, it cannot be a proper subinterval.
