By a real torus I mean the Lie group $T = \mathbb{R}^2/\langle e_1,e_2\rangle$ where $e_1 = (1,0)$ and $e_2 = (0,1)$.
Let $M_2(\mathbb{Z})^+$ denote the set of 2x2 integer matrices with positive (hence nonzero) determinant. Given $D\in M_2(\mathbb{Z})^+$ with determinant $n$, it induces a self-map of $T$ with kernel of order $n$.
Is there a nice characterization of the matrices in $M_2(\mathbb{Z})^+$ which have cyclic kernel?
If $D = D_n := [[1,0],[0,n]]$, then it's easy to check that: $$SL(2,\mathbb{Z})\cap D_n\cdot SL(2,\mathbb{Z})\cdot D_n^{-1} = \Gamma_0(n)$$
Now suppose $D$ is any matrix in $M_2(\mathbb{Z})^+$ with cyclic kernel of order $n$. Is the intersection $$SL(2,\mathbb{Z})\cap D\cdot SL(2,\mathbb{Z})\cdot D^{-1}$$ always conjugate to $\Gamma_0(n)$?
The matrix $A\in M_2(\Bbb Z)^+$ will have cyclic kernel iff the gcd of its entries is $1$.
This condition is necessary: if the gcd is $g>1$ then the $g$-torsion of $T$ is contained in the kernel, and the $g$-torsion is non-cyclic.
The condition is sufficient. Consider the Smith normal form $S$ of $A$. Then the kernel of $S$ will be isomorphic to that of $A$. But if the gcd is $1$, then $S=\pmatrix{1&0\\0&\det A}$ which has kernel cyclic of order $\det A$.
In higher dimensions this argument still works: $A\in M_n(\Bbb Z)^+$ will have cyclic kernel iff the Smith normal form has diagonal entries $1,\ldots,1,\det A$.