How to show mathematically that the electric field inside a conductor is zero?

The boundary conditions by themselves can't tell you anything about a conductor. The boundary conditions can't even tell which side of the surface has the conductor!

One way to model a conductor is as an Ohmic conductor where there is a constant $\sigma$ (different than the surface charge density listed in your boundary conditions) and then you assert the Ohmic condition:

$$\vec J=\sigma \vec E$$ and then you can take the divergence of both sides and get $$\sigma\frac{\rho}{\epsilon_0}=\sigma\vec \nabla \cdot \vec E = \vec \nabla \cdot \vec J$$

Where we used the Maxwell equation $\dfrac{\rho}{\epsilon_0}=\vec \nabla \cdot \vec E$ and we can also take the divergence of $$\vec \nabla \times \vec B=\mu_0\vec J+\mu_0\epsilon_0\frac{\partial \vec E}{\partial t}$$ to get the continuity equation

$$\vec \nabla \cdot \vec J=-\epsilon_0\vec \nabla \cdot\frac{\partial \vec E}{\partial t}=-\frac{\partial \rho}{\partial t}.$$

This means we have $$\frac{\partial \rho}{\partial t}=-\vec \nabla \cdot \vec J=-\frac{\sigma}{\epsilon_0}\rho.$$

So you might have started with an initial charge density but every place inside the conductor it exponentially decays over time.

And this relates to the initial value formulation of Electrodynamics. You start with an actual physical electromagnetic field at a moment and then it evolves according to $$\frac{\partial \vec B}{\partial t}=-\vec\nabla \times \vec E, \text{ and}$$

$$\frac{\partial \vec E}{\partial t}=\frac{1}{\epsilon_0}\left(\frac{1}{\mu_0}\vec\nabla \times \vec B-\vec J\right)$$

So the fields at a later time are a consequence of the fields at an earlier time (and the current) and the evolution equations above.

So for an Ohmic material we know $\vec J$ so we can evolve the fields because the evolution equations are just Maxwell solved for the time rates of change.

So you had an initial charge distribution and an initial electric field. They might have been zero, they might have been nonzero.

I've seen many "conceptual" arguments that if there was a field the charges would move and produce a field canceling this one out.

If you think of statics as the long time limit of dynamics then you don't have to go conceptual. An Ohmic material literally does have a nonzero current where there is a nonzero electric field. But that current causes the charge to go away, you can imagine places where the charge density is positive and negative initially and the initial electric field streamlines could connect some of those and/or it can connect those places to the surface. And since the current points the same way we can see that this exponentially decreasing charge density is becasue opposite charge densities are canceling each other as charge flows or the charge imbalance is moving to the surface, thus increasing the charge density of the surface over time.

The charge density on the surface can change in a different way than exponentially decreasing over time. Why? Because $\sigma$ (from the Ohmic condition) is not constant across the surface on the boundary of the conductor. In fact, the boundary of the conductor might have a vacuum with $\vec J=\vec 0$ on the other side.

Can we argue that the electric field is zero inside? Yes and no. On the one hand if we assert that part of statics is $\vec J=\vec 0$ then we have $\vec E=\vec 0$ right away. But that's more like assuming. But if you allow $\vec J\neq \vec 0$ then your conductor could have a zero charge density everywhere but have a steady current as long as the boundary of the conductor is supplied with the current it needs for that steady current.

It's totally a valid solution of Maxwell to have a cylindrical infinite wire pointing in the $\hat z$ direction with a uniform nonzero $\vec J$ pointing in the $\hat z$ direction inside the cylindrical infinite wire.

So whenever you have a counterexample you know you need to strengthening your hypothesis. That situation can have a static unchanging electric field, but it has a nonzero current.

You need to use Ohms law: $J = \sigma E$ which has to be added to Maxwell's equations as a bulk observation, as explained by this answer.

You can then conclude that the electric field is zero in a conductor for:

• perfect conductor where $\rho = 1/\sigma = 0$ and $J$ is finite
• static case where $J = 0$ and $\sigma$ is finite