= Solution
The <Lorentz reciprocal theorem for Stokes flow> states that two Stokes solutions $(\mathbf u^{(1)},\boldsymbol\sigma^{(1)})$ and $(\mathbf u^{(2)},\boldsymbol\sigma^{(2)})$ in the same fluid domain satisfy
$$
\int_{\partial\mathcal D}
\mathbf u^{(1)}\cdot\boldsymbol\sigma^{(2)}\mathbf n\,dS
=
\int_{\partial\mathcal D}
\mathbf u^{(2)}\cdot\boldsymbol\sigma^{(1)}\mathbf n\,dS.
$$
Apply it to the same rod translating with arbitrary velocities $\mathbf U_1$ and $\mathbf U_2$. The integral at infinity vanishes, and the rod-surface integrals give
$$
\mathbf U_1\cdot\mathbf F_2
=\mathbf U_2\cdot\mathbf F_1.
$$
Since $\mathbf F_i=-\mathbf R\mathbf U_i$,
$$
\mathbf U_1\cdot\mathbf R\mathbf U_2
=\mathbf U_2\cdot\mathbf R\mathbf U_1
$$
for every pair of vectors. Therefore
$$
\boxed{\mathbf R=\mathbf R^T}.
$$
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