Since $\nabla \times \nabla \phi=0$ $$\nabla \times \mathbf B = \nabla\times (\mu \nabla \phi)\\ =\mu \nabla \times \nabla \phi+\nabla\mu \times \nabla \phi\\ =\nabla\mu \times \nabla \phi$$
$$\nabla \cdot \mathbf B = \nabla\cdot(\mu \nabla \phi)\\ =\mu \nabla^2\phi+\nabla\mu \cdot \nabla \phi\\$$
If $\mu$ is constant in a region then there $\nabla \mu = 0$ and thus $\nabla \cdot \mathbf B =\mu \nabla^2\phi =0$ and we get the special case of Laplace's equation for the magnetic potential $\nabla ^2 \phi=0.$ Other wise, in general, the only thing we know is that $\mu \nabla^2\phi+\nabla\mu \cdot \nabla \phi=0.$
Even if $\mu$ is "piece-wise" constant, that is $\nabla \mu = 0$ in discrete regions, you have at the boundary of those regions jump conditions: the normal component $B_n$ of $\mathbf B$ is continuous and the tangential component $H_t$ of $\mathbf H$ is continuous. These continuity conditions can be interpreted as surface sources generating a non-zero field within the constant $\mu$ regions.