= Solution
Put $\mathbf r=(x,y,z)$ with $\mathbf e_3$ along $z$, and interpret the symmetry as <axisymmetry> of the whole <mass density>. A half-turn about $\mathbf e_3$ sends $(x,y,z)$ to $(-x,-y,z)$ and leaves the integration domain and <mass density> invariant. Therefore $\int\rho xz\,d^3r=\int\rho yz\,d^3r=0$. A quarter-turn sends $(x,y,z)$ to $(-y,x,z)$; it gives
$$
\int\rho xy\,d^3r=-\int\rho xy\,d^3r=0,\qquad
\int\rho x^2\,d^3r=\int\rho y^2\,d^3r.
$$
All off-diagonal entries of the <inertia tensor> consequently vanish. Writing $s^2=x^2+y^2$, the <principal moments of inertia> satisfy
$$
\boxed{I_1=I_2=\int_B\rho\left(\frac{s^2}{2}+z^2\right)\,d^3r,\qquad
I_3=\int_B\rho s^2\,d^3r.}
$$
Thus every <orthonormal basis> with third vector along the symmetry axis is a <principal body frame>. This is <axisymmetry of the inertia tensor>. In general there is no requirement that $I_3=I_1$.
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