Is $\mathbb{Q}$ the orbit of a rational function under iteration? In this previous post I asked for the smallest set of continuous real functions that could generate $\mathbb Q$ by iteration starting from $0$. Surprisingly one continuous function suffices.
In the question I gave the example of three rational functions that generate $\mathbb{Q}$, $f(x)=1/x$, $g(x)=x+1$ and  $h(x)=x-1$. It would be interesting to know if this is best possible and in particular whether one rational function can generate all of $\mathbb{Q}$:

Can $\mathbb{Q}$ be generated as the orbit of fewer than 3 rational functions?

The question Orbits of rational functions asks a more general question but I don't think  explicitly answers it for  $\mathbb{Q}$ itself.
 A: A rational function is as a self-map of $\mathbb P^1$. With that understanding, as was noted earlier, it is possible to generate all of the points $\mathbb P^1(\mathbb Q)$ by starting with the point $0$ and applying elements of the semi-group $\langle f_1,f_2\rangle$ generated by iteration using the two functions $f_1=x+1$ and $f_2=-1/x$. In this construction, both $f_1$ and $f_2$ are rational maps of degree $1$.
However, if one instead uses sets of rational maps $f(z)\in\mathbb Q(z)$ of degree at least $2$, then no finitely generated semi-group of such rational maps has an orbit that contains all of $\mathbb P^1(\mathbb Q)$, and indeed, any such orbit will be fairly sparse. Here's a quick proof (shown to be by Wade Hindes). Let $\mathcal F=\langle f_1,\ldots,f_r\rangle$, where $f_i\in\mathbb Q(z)$ has degree $d_i\ge2$. Then we have the height estimate
$$ h\bigl(f_i(P)\bigr) \ge d_i h(P) - C(f_i). $$
It follows that for each $i$,
$$
f_i\bigl(\mathbb P^1(\mathbb Q)\bigr) := \bigl\{ f_i(Q) : Q \in \mathbb P^1(\mathbb Q) \bigr\}
$$
has density $0$, where we use the height function to count points. But then for any starting point $P \in \mathbb P^1(\mathbb Q)$, the full orbit satisfies
$$ \mathcal F(P) := \bigl\{ f(P) : f\in\mathcal F\bigr\}
\subseteq \bigcup_{1\le i\le r} f_i\bigl(\mathbb P^1(\mathbb Q)\bigr).
$$
Thus the orbit $\mathcal F(P)$ is the union of finitely many sets of density $0$, so the orbit $\mathcal F(P)$ has density $0$.
A: As was mentioned in the comments by pregunton, it is possible to do using two rational functions. I claim it is not possible using just one. As Fedor Petrov suggests in another comment, this is because rational functions of degree higher than $1$ are never going to be surjective, which can be shown with help of Hilbert's irreducibility theorem. Indeed, take a rational function $\frac{f(x)}{g(x)}$ with coprime polynomials $f$, $g$ of which at least one has degree greater than $1$. The polynomial $h(x,t)=tf(x)-g(x)\in\mathbb Q[x,t]$ is irreducible then, so by Hilbert's theorem there are infinitely many values $q\in\mathbb Q$ for which $h(x,q)\in\mathbb Q[x]$ is irreducible. For all but one of these $q$, $h(x, q)$ will have degree $\max(\deg f,\deg g)>1$, so irreducibility implies it has no rational roots. Hence $q$ is not in the image of $\frac{f(x)}{g(x)}$.
The only case remaining is that of $\deg f,\deg g\leq 1$. In this case either $\deg g=1$ and the rational function has a rational pole, so its iteration can't go over all rationals, or else it is affine of the form $ax+b$ and it's easy to see explicitly its iterations do not cover all rationals.
