Number of orbits of the natural action of order preserving bijections of $\mathbb Q$ on $\mathbb Q^n$

Question

By an order preserving map $f : \mathbb Q \to \mathbb Q$, let us mean $ x > y \implies f(x) > f(y)$ . Let $Aut (\mathbb Q) =\{ f : \mathbb Q \to \mathbb Q : f $ is bijective and order preserving $\}$. Then $Aut (\mathbb Q)$ acts naturally on $\mathbb Q^n$ as $f . (a_1,...,a_n)=(f(a_1),...,f(a_n))$. Then is it true that the number of orbits under this action is finite ?

Answer 1

Yes it is true. One way to prove it is to prove that two $n$-uples $(a_1,\dots,a_n)$ and $(b_1,\dots,b_n)$ are in the same orbit if and only if they are arranged in the same order, i.e. if and only if $a_i < a_j\Leftrightarrow b_i< b_j$ and $a_i=a_j\Leftrightarrow b_i=b_j$. So the number of orbits will be equal to the number of possible orderings of your $n$ numbers. Such an ordering is completely described by the function that gives to every $a_i$ its rank, so the set of such ordering is in bijection with the set of functions $\{1,\cdots,n\}\to \{1,\cdots,n\}$ whose image is $\{1,\cdots, m\}$ for some $m$. In particular, there must be less than $n^n$ such orders.

It's pretty clear that two $n$-uples in the same orbit must have the same order; for the other direction, it is enough to prove that if $a_1<a_2<\cdots <a_m$ and $b_1<b_2<\cdots<b_m$ then there exist an order-preserving bijection $f:\mathbb{Q}\to \mathbb{Q}$ such that $f(a_i)=b_i$ for all $i$. But for that you can just take a piecewise-affine function between the points $(a_i,b_i)$ and $(a_{i+1},b_{i+1})$; since all the $a_i,b_i$ are rationals, every piece will be $\mathbb{Q}$-affine, and thus map rationals to rationals bijectively.