I'm arriving at the conclusion that "$\nabla \times \vec{E} = 0$ on the surface of an equipotential sphere ($E_\theta = E_\phi = 0$) (as the field must be normal to an equipotential/conductor) implies $E_r = constant$ on said sphere.
$\left( \nabla \times \vec{E} \right)_r = \frac{1}{r \sin \theta} \left[ \frac{ \partial}{\partial \theta} \left( \sin \theta {E_\phi} \right) - \frac{\partial E_\theta}{\partial \phi}\right] = 0 $ since $E_\phi = E_\theta = 0$
$\left( \nabla \times \vec{E} \right)_\theta = \frac{1}{r} \left[ \frac{1}{\sin \theta}\left( \frac{\partial E_r}{\partial \phi}\right)- \frac{\partial}{\partial r} \left( r E_\phi \right) \right] = 0$ implies $\frac{\partial E_r}{\partial \phi} = 0$ since $E_\phi = 0$
$\left( \nabla \times \vec{E} \right)_\phi = \frac{1}{r} \left[ \frac{\partial}{\partial r} \left( r E_\theta \right) - \frac{\partial E_r}{\partial \theta}\right] = 0 $ implies $\frac{\partial E_r}{\partial \theta} = 0$ since $E_\theta = 0$
So we have that "If the field is normal to a surface of a conducting sphere (it has to be) and $\nabla \times \vec{E} = 0$ along the surface of the sphere (again it has to be) then
$ \frac{\partial E_r}{\partial \theta} = \frac{\partial E_r}{\partial \phi} = 0$ implies $E_r = constant$ along the sphere.
This seems mathematically sensible but I can think of examples where $E_r \neq constant$ along the surface of a conducting sphere. Namely when a point charge is put outside a conducting sphere. (Suppose we place the origin at the center of the conducting sphere in the picture below)
As you can see the field lines are not equally spaced. However the field is perpendicular to the surface.
What is my mistake in reasoning? What have I forgotten physically or mathematically?
EDIT: I made the mistake of assuming because the function value is $0$ at a point that its derivative is also $0$. I probably want to incorporate $\nabla \cdot \vec{E} = 0$ along with $ \nabla \times \vec{E} = 0$ implies $\frac{\partial E_r}{\partial \theta} = \frac{\partial E_r}{\partial \phi} = 0$