= Solution
The possible ranks and the corresponding <homeomorphism> types are
$$
\boxed{n=0:\ S^3;\qquad n=1:\ S^2\times S^1;\qquad n=3:\ T^3.}
$$
There is no example for $n=2$ or $n\geq4$. The uniqueness here is topological, not uniqueness of a <Riemannian metric>.
To see the restrictions, use the <prime decomposition of a closed orientable three-manifold>. Its <fundamental group> is the <free product> of the <prime three-manifold> groups. A <free product> of two nontrivial groups is nonabelian, so at most one summand can have nontrivial <fundamental group>. The remaining <simply connected> summands are three-spheres by the <Poincaré conjecture>. If the nontrivial summand is not irreducible, it is $S^2\times S^1$, giving rank one. If it is irreducible and has infinite <fundamental group>, the <sphere theorem for three-manifolds> makes it <aspherical>. Its <fundamental group> is then a dimension-three <Poincare duality group>. But $\mathbb Z^n$ is a duality group of dimension $n$, as seen from the <torus> classifying space or its <Koszul resolution>, so $n=3$.
For rank three, the flat case of the <geometrization theorem> says that a closed irreducible orientable <three-manifold> with <fundamental group> $\mathbb Z^3$ admits a flat metric. Write it as $\mathbb R^3/\Gamma$. By the <Bieberbach theorem>, $\Gamma$ contains a full-rank translation lattice $\Lambda$. If $g=(R,v)\in\Gamma$ commutes with translation by each $\lambda\in\Lambda$, then $R\lambda=\lambda$ for every lattice vector, forcing $R=I$. Since $\Gamma$ itself is abelian, it consists entirely of translations. Its quotient is therefore the three-torus. Rank zero follows directly from the <Poincaré conjecture>, and rank one was already determined by the <prime decomposition of a closed orientable three-manifold>. Historically, before the Poincare theorem, the uniqueness conclusions were phrased allowing <connected sums> with <homotopy> three-spheres; the modern theorem removes that ambiguity.
The six flat types below mean the six \b[closed orientable] flat <three-manifold> types. There are infinitely many individual flat metrics and there are additional noncompact <flat Riemannian manifolds>. In each closed case the <universal cover> is <Euclidean space> and the <deck transformation group> is a torsion-free crystallographic group. The <holonomy groups of closed orientable flat three-manifolds> are $1,C_2,C_3,C_4,C_6,C_2\times C_2$.
The first is a <flat torus> $\mathbb R^3/\Lambda$, with trivial holonomy. For each cyclic case choose a planar lattice $L$ admitting the relevant rotation $R_m$, and generate the <deck transformation group> by translations in $L$ and the screw
$$
s_m(v,z)=(R_mv,z+1/m),\qquad m=2,3,4,6.
$$
Its $m$th power is unit translation along the axis. A nontrivial holonomy element has nonzero axial displacement modulo integers, so has no fixed point. All transformations preserve orientation. A compact prism <fundamental domain> gives compactness. Equivalently, the manifold is a <torus> <mapping torus> with finite-order <monodromy>. In planar lattice bases one may take
$$
R_2=-I,\quad
R_3=\begin{pmatrix}-1&-1\\1&0\end{pmatrix},\quad
R_4=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\quad
R_6=\begin{pmatrix}0&-1\\1&1\end{pmatrix}.
$$
The half-turn uses any planar lattice, the quarter-turn a square lattice, and the third- and sixth-turns a hexagonal lattice. These give respectively the <half-turn flat three-manifold>, <third-turn flat three-manifold>, <quarter-turn flat three-manifold>, and <sixth-turn flat three-manifold>. Their holonomies have orders $2,3,4,6$, and their first <Betti numbers> are all one, since the only fixed direction of holonomy is the screw axis.
The sixth type is the <Hantzsche-Wendt manifold>. A concrete group is generated by
$$
\alpha(x,y,z)=(x+1/2,-y,-z),\qquad
\beta(x,y,z)=(-x,y+1/2,-z+1/2).
$$
Their linear parts are commuting half-turns about perpendicular axes. Moreover $\alpha^2$, $\beta^2$ and $(\alpha\beta)^2$ are translations by $(1,0,0)$, $(0,1,0)$ and $(0,0,-1)$. Their translation subgroup is the unit cubic lattice and their linear quotient is $C_2\times C_2$. In each nontranslation coset, the displacement along the fixed axis is a half-integer, so the action is free. The quotient is compact and orientable. No vector is fixed by both linear parts, so its first <Betti number> is zero. Bieberbach classification gives precisely these six affine types; the explicit screw descriptions distinguish them without identifying arbitrary finite quotients with holonomy.
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