= Solution
Apply the <Lorentz reciprocal theorem for Stokes flow> to $\mathbf u_1$ and the test flow around a sphere rotating with arbitrary $\widehat{\boldsymbol\Omega}$. On $r=a$,
$$
\widehat{\mathbf u}=\widehat{\boldsymbol\Omega}\times\mathbf x,
\qquad
\widehat{\boldsymbol\sigma}\mathbf n=-3\mu\widehat{\boldsymbol\Omega}\times\mathbf n.
$$
The reciprocal integral containing $\widehat{\mathbf u}\cdot\boldsymbol\sigma_1\mathbf n$ vanishes by the first-order torque condition. Since $\partial_r\mathbf u_0=-2\boldsymbol\Omega_0\times\mathbf n$, the first-order boundary velocity is
$$
\mathbf u_1=\mathbf U_1+a\boldsymbol\Omega_1\times\mathbf n+3f\boldsymbol\Omega_0\times\mathbf n.
$$
The translational term integrates to zero. Therefore, for every $\widehat{\boldsymbol\Omega}$,
$$
\int_{r=a}\left(a\boldsymbol\Omega_1\times\mathbf n+3f\boldsymbol\Omega_0\times\mathbf n\right)
\cdot(\widehat{\boldsymbol\Omega}\times\mathbf n)\,dS=0.
$$
Now
$$
\int_{r=a}(\mathbf A\times\mathbf n)\cdot(\mathbf B\times\mathbf n)\,dS
=\frac{8\pi a^2}{3}\mathbf A\cdot\mathbf B.
$$
For $f=a(\mathbf n\cdot\mathbf D\mathbf n)$, tracelessness of $mathbf D$ and the stated fourth-moment identity give
$$
\int_{r=a}f(\boldsymbol\Omega_0\times\mathbf n)
\cdot(\widehat{\boldsymbol\Omega}\times\mathbf n)\,dS
=-\frac{8\pi a^3}{15}\widehat{\boldsymbol\Omega}\cdot\mathbf D\boldsymbol\Omega_0.
$$
It follows that
$$
\boxed{\boldsymbol\Omega_1=\frac35\mathbf D\boldsymbol\Omega_0},
\qquad \alpha=\frac35.
$$
A centered ellipsoid has inversion symmetry. An applied axial couple is unchanged under inversion, whereas a translational velocity is reversed, so uniqueness of <Stokes flow> forces
$$
\boxed{\mathbf U_1=0}.
$$
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