Is there some $f_{2,3}:\Bbb R^+\to S^1$ such that $f_{2,3}(x)=f_{2,3}(2x)=f_{2,3}(3x)$?

Question

Let $f_2(x)=x^{2\pi i/\log2}$

Then I think we have that $f_2(x)=f_2(2x)$

And if we set $f_3(x)=x^{2\pi i/\log3}$

then we have that $f_3(3x)=f_3(x)$

Is there some way to combine these so that $f_{2,3}(x)=f_{2,3}(2x)=f_{2,3}(3x)$ but other natural number inputs are distinct?

(Of course this is a symmetric transitive and reflexive equivalence relation so when I say other natural numbers are distinct, I mean all 5-rough positive integers give distinct results.)

Answer 1

If you have a continuous $f : \mathbb R_{>0} \to \mathbb C$ with the property that $f(2x) = f(3x) = f(x)$ for all $x$, then $f$ is constant. This is because the set $2^{\mathbb Z} 3^{\mathbb Z}$ is dense in $\mathbb R_{>0}$. (Which is because $\log 3 / \log 2$ is irrational.)