= Solution
Now take zero <torque> and $\boldsymbol\omega(0)=(1,0,1)$. The axial <angular velocity> is $\omega_3=1$ and the complex transverse component obeys $\dot z=i\nu z$ with $z(0)=1$ and $\nu=(I_3-I_1)/I_1$. Hence the <angular momentum> components in the <principal body frame> are
$$
\boxed{\mathbf L_S(t)=\left(I_1\cos\nu t,\ I_1\sin\nu t,\ I_3\right),\qquad
\nu=\frac{I_3-I_1}{I_1}.}
$$
The transverse projection therefore traces a <circle> with signed <angular frequency> $\nu$. Its magnitude is constant, $L=\sqrt{I_1^2+I_3^2}$.
In the <inertial frame>, zero <torque> means that $\mathbf L$ is fixed. Using
$$
\mathbf L=I_1\boldsymbol\omega+(I_3-I_1)\omega_3\mathbf e_3
$$
and the <derivative of a body-fixed basis vector>, the symmetry axis satisfies
$$
\dot{\mathbf e}_3=\boldsymbol\omega\times\mathbf e_3
=\frac{\mathbf L}{I_1}\times\mathbf e_3.
$$
Thus there is uniform <torque-free precession of a symmetric top> about the fixed <angular momentum>, at a constant cone <angle> $\beta$ and space <angular frequency>
$$
\boxed{\cos\beta=\frac{I_3}{\sqrt{I_1^2+I_3^2}},\qquad
\Omega_{\mathrm{space}}=\frac{\sqrt{I_1^2+I_3^2}}{I_1}.}
$$
The cone <angle> stays constant, with no <nutation>. The <body-frame and space-frame frequencies of free symmetric-top precession> differ: $\nu$ describes rotating components, whereas $\Omega_{\mathrm{space}}$ describes the symmetry axis in space. The formulas also cover $I_3=I_1$, where the <angular momentum> components in the <principal body frame> are constant.
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