The group $\mathrm{PSL}_2(\mathbf{Q})$ of fractional linear transformations $x \mapsto (ax+b)/(cx+d)$ such that $a,b,c,d \in \mathbf{Q}$ and $ad-bc = 1$ acts on $\mathbf{Q} \cup \lbrace \infty\rbrace$. The weakest version of my question is the following:
Is there a short proof that no element of $\mathrm{PSL}_2(\mathbf{Q})$ acts transitively on $\mathbf{Q} \cup \lbrace \infty\rbrace$?
It seems obvious to me that no element of $\mathrm{PSL}_2(\mathbf{Q})$ can have finitely many orbits on $\mathbf{Q} \cup \lbrace\infty\rbrace$, but I have not been able to prove this.
Is this stronger claim true, and if so, does it have a short proof?
Some remarks and motivation: (1) The subgroup of $\mathrm{PSL}_2(\mathbf{Q})$ generated by the maps $x \mapsto x+1$ and $x \mapsto x/(x+1)$ acts transitively on $\mathbf{Q} \cup \lbrace \infty\rbrace$; these elements are used to generate the Calkin–Wilf tree of rational numbers.
(2) By conjugating by elements of $\mathrm{PGL}_2(\mathbf{Q})$ it suffices to look at maps of the form $g : x \mapsto 1/(b-x)$. If $|b| > 2$ then $g$ is conjugate in $\mathrm{PGL}_2(\mathbf{R})$ to a map of the form $x \mapsto \lambda x$. It follows that the iterates of $g$, starting at the initial value $x_0$ are of the form
$$ \frac{1}{b-\frac{1}{b-\ldots\frac{1}{b-x_0}}} $$
and converge to the continued fraction $1/(b-1/(b- \ldots ))$. The iterates of $g^{-1}$ behave similarly. So in this case each orbit of $\langle g \rangle \le \mathrm{PSL}_2(\mathbf{Q})$ has precisely two limit points in $\mathbf{R}$, namely the two roots of $x^2-bx+1$. Clearly this implies that $\langle g \rangle$ has infinitely many orbits on $\mathbf{Q} \cup \lbrace \infty\rbrace$. However if $|b| < 2$ then I have not been able to make this approach work.
(3) If $\mathbf{Q}$ is replaced with the finite field $\mathbf{F}_p$ then is not too hard to show that no element of $\mathrm{PSL}_2(\mathbf{F}_p)$ acts transitively on $\mathbf{F}_p \cup \lbrace \infty \rbrace$: see Lemma 8.2 in this paper with John Britnell. This leads to a proof of the result for $\mathbf{Q}$ by reduction mod $p$.