How to rule out counter-examples like Minkowski's question mark function

Question

Let $f$ be a function such that

1. $f:[0,1]\cap\mathbb Q\to\mathbb R$,
2. $f(0)=0$, and
3. $f'(x)=0$ for all $x$ in the domain.

Surprisingly, these conditions do not determine a unique function. The zero function satisfies them but so does Minkowski's question mark function. I'm interested in what can be added as a fourth condition to ensure that only the "natural option", i.e. in this case the zero function, is picked out. The condition should also work when condition 3 is replaced by $f'$ being any other uniformly continuous function.

Answer 1

I will start out the list myself:

4. $f$ has the Fundamental Theorem of Calculus-Property

By this I mean that for all $x$ in the domain it is the case that $f(x)=\int_0^xf'(t)dt$, where the integral is the Riemann integral.