Group of order-automorphisms of the rational numbers

Question

What reference do you recommend for the group $\mathrm{Aut}(\mathbb{Q},<)$ of all order-automorphisms of the rational numbers? Needless to say, this is not about field-automorphisms.

Obviously, each partition of $\mathbb{Q}$ into countably-many nonempty intervals $I_i$, and any choice of an increasing bijection $I_i\rightarrow I_i$ for each $i$ yields such an automorphism, which already implies that there are at least $2^{\aleph_0}$ order-automorphisms of $\mathbb{Q}$. I would like to read a modern comprehensive treatment of the group $\mathrm{Aut}(\mathbb{Q},<)$.

It appears to be known that it is Polish, for example, for which I cannot find a complete proof.

I am aware of Discussion on order-automorphisms of the rational numbers.

Answer 1

Not an answer, just a proposition about how to construct those order-preserving bijections :

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Given a bijection $h : \mathbb{N} \to \mathbb{Q}$ and an order-preserving bijection $f : \mathbb{Q} \to \mathbb{Q}$, call $\text{recursion}(f,f^{-1},\, (-\infty,+\infty)\, )$

 recursion (order-preserving bijection : F, G, 
            interval of the rationals : (a,b)   ) 
 {
     i := 0
     loop { 
       if ( h(i) in (a,b) ) break
       i := i+1
     }
     c := h(i)
     recursion(G,F, (F(a),F(c)) )
     recursion(G,F, (F(c),F(b)) ) // F and G are switched at each call
 }

At each call of $\text{recursion}$ all we know is that $F$ is an order-preserving bijection $(a,b) \to (F(a),F(b))$.

Thanks to the bijection $h$, for any $x \in \mathbb{Q}$ the function $\text{recursion}$ will be called with $a = x, F= f$ and $a=x,F = f^{-1}$ after a finite time.

Claim : we can modify this recursion for constructing any order-preserving bijection, simply by choosing for $F(c)$ the value we'd like $ \in (F(a),G(b))$.

By construction, $f$ and $f^{-1}$ are order-preserving functions $\mathbb{Q} \to \mathbb{Q}$, hence $f$ is an order-preserving bijection $\mathbb{Q} \to \mathbb{Q}$.

And by construction again, given any order-preserving bijection $F$, the algorithm can reproduce $f = F$ by choosing $f(c) = F(c)$.

Given $F_1,F_2$ two order-preserving bijections, we can easily reproduce $f = F_1 \circ F_2$.