Let $(\Omega,\leq)$ be a dense linear order without endpoints. If $\Omega$ is countable, we know by Cantor that $(\Omega,\leq)$ is order-isomorphic to $(\mathbb{Q},\leq)$.
Suppose that $\Omega$ is not countable. In case of $\mathbb{Q}$, we know that for any two ordered sequences $s$ and $\widetilde{s}$ in $\mathbb{Q}$, there is an order-isomorphism which sends $s$ to $\widetilde{s}$. (Call this property (IND).)
Is this property also true for uncountable $\Omega$ ?
Secondly, Cantor used a Back-and-Forth argument to prove the aforementioned result. Is there an uncountable variation of this technique (for instance, using transfinite induction) ?
As Noah's comment points out, the Cantor theorem does not generalize quite so directly.
However, Hausdorff found a generalization in his notion of an $\eta$-set (see here). $(R,<)$ is an $\eta_\alpha$ set if $|R|=\aleph_\alpha$, and given any two subsets $X<Y$ of $R$, each of cardinality less than $\aleph_\alpha$, there is a $u\in R$ with $X<u<Y$. (By $X<Y$, I mean all elements of $X$ are less than all elements of $Y$).
The generalization of the Cantor result reads: if $A,B\subset R$ with $|A|=|B|<\aleph_\alpha$, and there is an order-preserving map $A\to B$, then this map can be extended to an order-preserving bijection of $R$ to itself. This is proved by a transfinite back-and-forth argument, just like you suggested.
Nowadays Hausdorff's $\eta$ sets are considered one of the earliest examples of saturated structures.
Incidentally, the back-and-forth argument isn't due to Cantor: he proved his theorem with a one-way argument. It was that same Hausdorff who invented the back-and-forth method.