I would like to construct a (conformal) map similar to the exp conformal map, but for which the vertical lines in the source space are evenly separated in the target space. That is, in this simulator: https://www.geogebra.org/m/qCSzkRsP (have to change the function to exp(z)), I would like the red lines to be evenly spaced in the target space.
In the end, what I really want is the inverse transform of that (to be able to "unfold" an 2D object), so it will likely be a modified log conformal map. But I find it easier to imagine it with the exp conformal map first.
Note: I do care about the fact that there is a notch around the origin in the target space.
There's no such conformal map.
Suppose we had such a map, $f(x, y) = \rho(x, y) e^{i \alpha(x, y)}$. The condition that vertical lines in the domain are mapped to circles centered at the origin is $$\rho_y = 0 ,$$ and given that it is satisfied, the condition that evenly spaced vertical lines in the domain are mapped to evenly spaced circles is then $$\rho_x = C,$$ for some constant $C$, that is, $\rho_{xx} = \rho_{xy} = 0$.
In these coordinates, the Cauchy-Riemann equations are $$(\log \rho)_x = \alpha_y, \qquad (\log \rho)_y = -\alpha_x.$$
Substituting our conditions gives that $$ \alpha_x = 0, \qquad \alpha_y = \frac{C}{\rho} . $$ But these conditions imply that $\rho$, $\alpha$ are constant. (Why?) Thus, $f(x, y) = \rho e^{i \alpha}$ is constant, hence not conformal.
Dropping the conformality hypothesis In the comments o.p. mentioned they were also in interested a function that had the desired spacing property even without the conformality requirement. In that case, applying our conditions on $\rho$ imposes precisely that $\rho$ is an affine function of $x$ alone, that is, is of the function $$f : (x,y) \mapsto (A x + B) e^{ig(x,y)} .$$ Here, $A$ is the distance between the concentric rings that are the images of vertical lines $1$ unit apart, and $B$ is the radius of the circle to which the imaginary axis is mapped. In particular, we probably want $A \neq 0$, lest our function map all of $\Bbb C$ into the circle of radius $B$ centered at the origin). The Jacobian determinant of $f$ as above is $$\det \operatorname{Jac}(f) = A (A x + B) g_y,$$ so if $A \neq 0$ and $g_y$ vanishes nowhere, the map is invertible everywhere except on the vertical line $x = -\frac{B}{A}$, which is mapped to the origin. The natural choice $B = 0$, $g(x,y) = y$ yields the simple map $$(x,y) \mapsto x e^{i y},$$ perhaps restricted to the right half-plane, $\{x > 0\}$.
Remark 2 If we only impose the condition $\rho_y = 0$, an analysis similar to ours shows that the only conformal maps that send vertical lines in the domain to circles centered at the origin are (locally) the maps $z \mapsto A \exp(b z)$.