Consider a solid sphere of radius $R$ centered at the origin. I know that since there is rotational symmetry about any axis $\hat{n}$ we have that the diagonal elements of the inertia tensor will all be equal. It isn't clear to me why the off-diagonal elements are zero though.
I can compute them explicitly by converting to spherical coordinates and actually working it out, but is there any way to see why this should be true without having to compute it?
On
You can write the $xy$ component of the tensor as $I_{xy} = -\int_V xy dm=-\int_V xy \rho dxdydz$. We know that the surface of the sphere is given by $x^2+y^2+z^2=R^2$, so we can write out the tensor component as a triple integral $$I_{xy} = -\rho \int_{-R}^{R} dz\int_{-R^2+z^2}^{R^2-z^2}y dy \int_{-R^2+z^2+y^2}^{R^2-z^2-y^2}xdx$$ You can notice that in the last integral the integration occurs symmetrically around 0, and the integrand is an odd function, so the integral is 0.
On
Let your body be $\Omega$ with uniform density $\rho$ and we want to find the off-diagonal elements via a usual expression
$$I_{xy} = -\rho \int_\Omega xy \,dx\, dy \,dz .$$
Suppose now that your body $\Omega$ be symmetric with respect the plane $yOz$, then we can make an orthogonal change of variables $(x,y,z)\to (-x,y,z)$: $$I_{xy} = -\rho \int_\Omega (-x)y \,dx\, dy \,dz = -I_{xy},$$ hence $I_{xy}=0$.
Since the solid sphere centered in origin is symmetric with respect to all planes passing through the origin, you can conclude that all off-diagonal entries are zero.
The same argument, for example, allows you to say that an axis-aligned cuboid centered in origin also has zero off-diagonal elements in its inertia tensor.
By symmetry the angular momentum is proportional to angular velocity. The only way to have this is when off diagonal elements of the inertia tensor are null.