= Solution
Relative positions and velocities measured from galaxy $A$ are $r_B-r_A$ and $v_B-v_A$. <Spatial homogeneity> says that the same position-to-relative-velocity rule applies from every galaxy. Thus $v(r_B-r_A)=v(r_B-r_O)-v(r_A-r_O)$. Taking $r_O=0$ and putting $r_B=x+y$, $r_A=y$ gives additivity $v(x+y)=v(x)+v(y)$. For a continuous physical velocity field, additivity implies rational homogeneity and, by continuity, real homogeneity. Therefore
$$
\boxed{v(r)=Hr}
$$
for a position-independent <matrix> $H$. Regularity is necessary to exclude pathological discontinuous additive functions.
Decompose the given <matrix> as $H=(5D/t)I+(D/t)W$, where
$$
W=\begin{pmatrix}0&-1&-2\\1&0&-1\\2&1&0\end{pmatrix}.
$$
Its symmetric part is isotropic: no <cosmological shear>, expansion scalar $\nabla\cdot v=15D/t$, and linear scale factor proportional to $t^{5D}$ for $t>0$. Its antisymmetric part is rigid rotation with <angular velocity>
$$
\boxed{\boldsymbol\omega(t)=\frac D t(1,-2,1),\qquad|\boldsymbol\omega|=\frac{\sqrt6D}{t}}.
$$
The <vorticity> is $2\boldsymbol\omega$. Comoving separations obey
$$
r(t)=\left(\frac t{t_0}\right)^{5D}
\exp\!\left[D\log(t/t_0)W\right]r(t_0).
$$
Thus all distances grow by the same power, volumes grow as $t^{15D}$, and the pattern rotates about the fixed axis $(1,-2,1)$ through angle $\sqrt6D\log(t/t_0)$. Homogeneity alone permits this rotation; full cosmological isotropy would not.
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