= Solution
For a uniform circular cross-section, averaging gives $\langle x^2\rangle=\langle y^2\rangle=R^2/4$; for a uniform axial coordinate in $[-h,h]$, $\langle z^2\rangle=h^2/3$. Symmetry makes the coordinate axes principal axes, since all products of distinct coordinates integrate to zero. Thus the <principal moments of inertia> about the centre are
$$
\boxed{I_x=I_y=M\left(\frac{R^2}{4}+\frac{h^2}{3}\right),\qquad I_z=\frac{MR^2}{2}}.
$$
The <inertia tensor> is a scalar multiple of identity exactly when $h^2/3=R^2/4$. Therefore
$$
\boxed{\frac{2h}{R}=\sqrt3}
$$
is the height-to-radius ratio for sphere-like rotational inertia; the height-to-diameter ratio is $\sqrt3/2$. The geometric shape remains cylindrical, but its angular momentum is parallel to angular velocity for every axis, just as for a sphere.
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