Are the hyper-reals countably transitive? A hyper-real field is $ R^*=(R^N)_{/U}$ where $U$ is a free ultrafilter on $N$. 
If A and B are any countable order-isomorphic subsets of $R^*$, is there an order-automorphism of $R^*$ that maps $A$ onto $B$ ?
(This is a follow-up to my question Is there a non-trivial countably transitive linear order?)
 A: This is independent of ZFC. If the continuum hypothesis holds, the answer is yes, but it is also consistent that the answer is no.
To explain further, I'll have to use some model-theory terminology. If you're not familiar with it, you might want to take a look at this wikipedia page.
Countable transitivity is called "strong $\aleph_1$-homogeneity" in model theory. In general, a structure is strongly $\kappa$-homogeneous if any isomorphism between substructures of size $<\kappa$ extends to an automorphism. 
Why does CH imply that $\mathbb{R}^\mathbb{N}/U$ is strongly $\aleph_1$-homogeneous? Well, any nonprincipal ultraproduct is $\aleph_1$-saturated, and if CH is true, then $|\mathbb{R}^\mathbb{N}/U| = \aleph_1$ (this is because $\mathbb{R}^\mathbb{N}/U$ is a quotient of a set of size $\aleph_1^{\aleph_0} = 2^{\aleph_0} = \aleph_1$, and it can't be countable - $\mathbb{R}$ embeds in it). So $\mathbb{R}^\mathbb{N}/U$ is an $\aleph_1$-saturated structure of size $\aleph_1$. In this case we just say it's saturated. 
Now saturated models are strongly homogeneous. I.e. if a structure of size $\kappa$ is $\kappa$-saturated, then it's $\kappa$-strongly homogeneous. This is exactly what we want (with $\kappa = \aleph_1$). You can prove this easily by a back and forth argument: the saturation allows you to keep extending your automorphism by realizing types over what you have so far.
