I am asked if such a conformal map exists. My strategy has been to find an isothermal parametrization of the helicoid. Moreover I know that we must have $E=G=\phi$, $F=0$ for some $\phi$ and by virtue of Gauss's Egregium Theorem, $\phi$ must be constant. I have tried changing $u$ by some $f(u)$ as Do Carmo sometimes does but have reached nowhere. Is there a simple argument to show such a map does not exists or to construct such parametrization?
Could I use that in such case that the plane has Gaussian curvature $0$ while the Helicoid in such parametrization verifies: $$ K = (EG-F^2)/(eg-f^2)= \phi^2/(eg-f^2)\neq0$$
Thank's in advance.
There does exist such a (local) conformal map. I'll present a very brief approach which uses the uniformization theorem, as Moishe Kohan suggested. The helicoid is simply connected (because it deformation retracts onto a helix which deformation retracts onto a point), therefore (by the uniformization theorem) it is conformally equivalent to either $\mathbb{H}^2, \mathbb{R}^2$ or $\mathbb{S}^2$. All of the former are (locally) conformally flat.
I can't think of any explicit example of such a map right now, however.