In trying to see how Banach's fixed point theorem would break down in an incomplete space, I tried to come up with an example of a function: $f: \mathbb{Q} \longrightarrow \mathbb{Q} \ \ $such that $ \forall x,y \in \mathbb{Q}$ our function is contracted with factor $\frac12$ and has no fixed points. i.e. $$ \left | {f(x)-f(y)} \right |\leq \frac12 \left | x-y \right | ,\\ \nexists \ \ x^* \in \mathbb{Q} \ \ s.t. \ f(x^*)=x^* \ $$ Starting with $f(x)=x^2-2$ and using Newton-Raphson I was able to construct the following function which works $$ f(x) := \left\{\begin{array}{lr} \frac12 (x-1) + \frac32, & \text{for } x < 1\\ \frac12- x^{-2}, & \text{for } 1\leq x\leq 2\\ \frac12 (x-2) + \frac32, & \text{for } x>2 \end{array}\right\}$$
That led me to thinking if there was a bijection which satisfied the same criteria. I think I can see why one could exist but am not sure how to prove that one necessarily exists. I am also keen on an explicit example of a bijection that works.
Consider for example $$ f(x) = \begin{cases} x/2 + 3/2 & x\le 2 \\ 3-1/x & 2 \le x \le 4 \\ x/2 + 3/4 & x \ge 4 \end{cases} $$ In the reals the only fixpoint is at $x=\frac{3+\sqrt5}2 \approx 2.6 $, which is irrational.
Because the curved segment in the middle uses $x^{-1}$ instead of $x^{-2}$, its inverse is a rational function, so the full $f$ is a contractive bijection $\mathbb Q\to\mathbb Q$.