I understand that any charge distribution in a conductor will always manifest itself onto its surface, and this leads to $\vec \nabla\cdot \vec E = 0$. But $\vec \nabla\cdot \vec E = 0$ doesn't mean that $\vec E$ has to be zero at all. In fact, coupled with the $\vec \nabla \times \vec E = 0$, the governing equation becomes Laplace's equation for which $\vec E = 0$ (or $V =$ constant) is just one solution out of a family of solutions. Is it possible, due to different boundary conditions, that the field in a conductor be any of those other solutions than simply just $\vec E = 0$?
The most given explanation is that any non-zero field in a conductor would immediately produce a non-steady current to dissipate that field until electrostatic equilibrium. But that assumes $\vec J$ to be non-steady, which isn't always the case (e.g. a conducting wire with constant current flow. Zero net internal charge but $\vec E$ is non-zero and $\vec J$ is constant). Could there be any other such steady $\vec J$ that arises electrostatically in conductors in real life?