Yes. This answer is based on the answers to your previous question.
Start with a computable ergodic map $T$ (D. Thomine constructs an example here). For every basic open neighborhood $B_i$, the set $D_i = \bigcup_n T^{-n}(B_i)$ is a dense effectively open set, uniformly in $i$. So some computable real $p$ meets all the $D_i$, meaning $X = \{ T^n(p) : n \in \omega\}$ is dense.
There is a computable, order-preserving bijection $g: X \to \mathbb{Q}$ (back-and-forth argument), and we may assume that $g(p) = 0$. $g$ induces a (bi-computable) homeomorphism $G: \mathbb{R} \to \mathbb{R}$.
Define $f = G\circ T\circ G^{-1}$.
Constructing the bijection:
We'll build a computable sequence of rationals $(q_n)_{n \in \omega}$ and define $g(T^n(p)) = q_n$.
Begin with $q_0 = 0$. Then compute enough of $p$ and $T(p)$ to determine how they are ordered ($p < T(p)$ or $T(p) < p$). If $p < T(p)$, choose $q_1$ to be a rational greater than 0; otherwise, choose $q_1$ to be a rational less than 0. In either case, choose $q_1$ to be the appropriate rational with the smallest Gödel number.
Then compute enough of $p$, $T(p)$ and $T^2(p)$ to determine how they are ordered, and choose $q_2$ to be a rational in the same relative position to $q_0$ and $q_1$. Again, choose the appropriate rational of smallest Gödel number.
Etc.
This is all a computable process, so $g$ is computable. It's a total order-preserving injection by construction.
Surjectivity is by induction on Gödel number. For a rational $r$, by the inductive hypothesis all rationals of smaller Gödel number are in the range of $g$. So fix an $n$ such that all rationals of smaller Gödel number occur in $q_0, \dots, q_{n-1}$, and assume $r$ does not occur in this list, as otherwise we are done. Define $C = \{i < n : q_i < r\}$. By the density of $X$, there is an $m \ge n$ with $T^i(p) < T^m(p)$ for all $i \in C$, and $T^m(p) < T^i(p)$ for all $i < n$ with $i \not \in C$. Fix the least such $m$. Then by construction, $q_m = r$.
