I Charge on conductors

The reasoning is very simple. I don't know, why Griffiths likes to obscure things in his textbooks. In electrostatics by definition we have all fields and sources time-independent and also no current density (see #3). Then within a conductor you have
$$\sigma \vec{E}=\vec{j}=0 \; \Rightarrow \; \vec{E}=0.$$
The electric charge density at every point is given by Gauss's Law (Heaviside-Lorentz units)
$$\vec{\nabla} \cdot \vec{E}=\rho.$$
Since ##\vec{E}=0## inside the conductor also ##\rho=\vec{\nabla} \cdot \vec{E}=0## inside the conductor. So if the conducting body is overall charged, the charge must be on its surface.

Caveat: To be more accurate, of course microscopically the entire body consists of charged particles (atomic nuclei and electrons). So there are charges inside the body but they are compensating each other precisely in the static case, when looked from a macroscopic (coarse-grained) point of view. So within macroscopic electrodynamics the macroscopic charge distribution vanishes in conductors for electrostatic situations.

 

Yes. The surface is equipotential.

That does not mean that the surface charge is uniformly distributed.

 

I didn't mention current, nor did I imply it.

The surface charge density is not one uniform value over the whole surface. It varies from location to location, but it is static.