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 $0$.

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