Note that element in Heisenberg group $\mathbb{H}^3$, whose entries are integers, has the form $$ \left( \begin{array}{ccc} 1 & a_1 & a_3 \\ 0 & 1 & a_2 \\ 0 & 0 & 1 \\ \end{array} \right),\ a_i\in \mathbb{Z} $$
If $A_i$ is set of such elements s.t. $a_{i+1}=a_{i+2}=0$, then $A_3$ is a center in a group $$G:=A_1\cup A_2\cup A_3,$$ which has an left isometric action on $\mathbb{H}^3$.
So $G/A_3$ acts on $X:=\mathbb{H}^3/A_3$. Here $$\left(\begin{array}{ccc} 1 & 0 & a_3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)\left(\begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \\ \end{array} \right) =\left(\begin{array}{ccc} 1 & x & z+a_3 \\ 0 & 1 & y \\ 0 & 0 & 1 \\ \end{array} \right)$$
Hence $X$ is homeomorphic to an open solid torus. In further note that $X$ is not a product or space form.
Question : Does $X$ have no conjugate point ?
Here $\mathbb{H}^3$ has a Riemannian metric $$ g=dx^2 + dz^2 + (x^2+1)dy^2-x\{ dydz +dzdy\} $$ where $\left(\begin{array}{ccc} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \\ \end{array} \right)$ is a coordinate parametrization for $\mathbb{H}^3$.
Here $c_1(t)=(0,0,t),\ c_2(t)=(x,y,z+t)$ are two shortest geodesics s.t. they connect two points in orbits $A_3\cdot (0,0,0),\ A_3\cdot (x,y,z)$, respectively. And they have same lengths.
So some closed smooth geodesic $c$ in $X$ has closed smooth geodesics around $c$ of same lengths. Hence the question is true ?
[Known result]
(1) $n$-dimensional torus with no conjugate point is flat.
(2) ${\rm SL}(2,\mathbb{R})$ that is homeomorphic to an open solid torus has a conjugate point.
Thank you for your attention.