= Solution
Let the unit normal $\mathbf n$ point from material 1 to material 2, and let their relative <magnetic permeabilities> be $\mu_1,\mu_2$, so $\mathbf B_i=\mu_0\mu_i\mathbf H_i$ for linear isotropic media. Apply <Gauss's law for magnetism> to an infinitesimal <Gaussian pillbox> straddling the interface. The side contribution vanishes as its height tends to zero, leaving
$$
\boxed{\mathbf n\cdot(\mathbf B_2-\mathbf B_1)=0.}
$$
The normal <magnetic flux density> is continuous.
Next use a narrow rectangular <Amperian loop> crossing the interface. The two short-side contributions vanish in the zero-height limit. The <magnetostatic Ampère law> for <magnetic field intensity> gives
$$
\boxed{\mathbf n\times(\mathbf H_2-\mathbf H_1)=\mathbf K_f,}
$$
where $\mathbf K_f$ is any free <surface current density>. Thus, if there is no free <surface current density> on the interface, the tangential <magnetic field intensity> is continuous. In that case the component relations are
$$
\boxed{B_{1n}=B_{2n},\qquad H_{1t}=H_{2t},\qquad
\mu_1H_{1n}=\mu_2H_{2n},\qquad
\frac{B_{1t}}{\mu_1}=\frac{B_{2t}}{\mu_2}.}
$$
The distinction between free and bound <electric currents> matters: bound currents due to <magnetization> are already incorporated in the definition of <magnetic field intensity>. A change of <magnetic permeability> alone does not make the tangential <magnetic flux density> continuous.
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