Is the Fabius function the unique solution (up to scaling) of the following functional differential equation?
$$ f'(x) = 2 f(2 x) $$
If so, how can this be proven? As I understand it, the Picard–Lindelöf theorem cannot be applied here, since the equation is not of the correct form.
If not, what minimal additional conditions are necessary to fix the Fabius function uniquely (up to scaling)?
The most upvoted answer to the question How to compute the values of this function ? ( Fabius function ) gives a way to compute the value of $f$ at any rational of the form $a/2^n$, using uniquely $$f(0) = 0 \tag{1}$$ $$f(1) = 1 \tag{2}$$ $$f'(x)=\begin{cases} 2f(2x) & 0\leq x\leq 1/2 \\ 2f(2(1-x)) & 1/2\leq x\leq1 \end{cases} \tag{3} $$ Because $f$ is differentiable, it is continuous, and the rationals $a/2^n$ are dense in $[0,1]$, therefore $f$ is uniquely determined by these three properties.
By replacing $f$ by $rf$ for any real $r \in \mathbb{R} \setminus \{1\}$, you obtain a function which verifies conditions 1 and 3, but not 2, and which is different from $f$, so condition 1 and 3 are not sufficient to determine $f$ uniquely (you could even take $f \equiv 0$). I don't know what happens when we take only conditions 2 and 3.