A linear order $<$ on a set $S$ is countably transitive iff, whenever $A$ and $B$ are order-isomorphic countable subsets of $S,$ there is an order-automorphism of $S$ which maps $A$ onto $B.$ Does such an order exist on some infinite set?
One book defined $n$-transitive for finite $n>0$ to mean that whenever $A$ and $B$ are $n$-element subsets, there is an order-automorphism that maps $A$ onto $B.$ It is easy to show that $2$-transitive implies $n$-transitive for all finite $n>2.$ I thought of extending this. I easily found an infinite $S$ ,with no largest member, where, whenever $A$ and $B$ are subsets of $S$, each order-isomorphic to $\omega$, there is an order-aut of $S$ that maps $A$ onto $B$ . Then I tried to extend it to this question. A solution must have the following properties : It is order-dense in itself, as no endpoints, has no $(\omega,\omega^*)$ gaps, and every countable subset is bounded and closed in the order topology.