Consider two (primitive) lattices $\pi_{i} \in \mathbb{C}$, such that $\pi_{1} = M \pi_{2}$ for $M \in \mathcal{S}_{m}$ with $$\mathcal{S}_{m}:=\Big\{\begin{pmatrix} A & B \\ 0 & D \end{pmatrix} \in \mathrm{GL}_{2}(\mathbb{Z}) : AD = m >0,\;D>0,\;0 \leq B \leq D\Big\}.$$ Does this imply that the elliptic curves that correspond to these lattices are isogenous and therefore one lattice corresponds to an order that it is an over-order (or the same order if the lattices are homothetic) to the order of the other lattice?

Consider two (primitive) elements $\pi_{i} \in \mathbb{C}$, such that $\pi_{1} = M \pi_{2}$ for $M \in \mathcal{S}_{m}$ with $$\mathcal{S}_{m}:=\Big\{\begin{pmatrix} A & B \\ 0 & D \end{pmatrix} \in \mathrm{GL}_{2}(\mathbb{Z}) : AD = m >0,\; 0 \leq B \leq D\Big\}.$$ Let $\Lambda_{i} = \mathbb{Z}\pi_{i} + \mathbb{Z}$ denote the corresponding lattice. Does this imply that: (a) the elliptic curves that correspond to these lattices are isogenous, and  (b) the conductor of the order that corresponds to $\pi_{1}$ divides the conductor of the order that corresponds to $\pi_{2}$, which equals det($M$).

If two (primitive) lattices $\pi_{i} \in \mathbb{C}$ are such that $\pi_{1} = M \pi_{2}$ for $M \in \mathcal{S}_{m}:=\{\begin{pmatrix} A & B \\ 0 & D \end{pmatrix} \in GL_{2}(\mathbb{Z}) : AD = m >0, D>0, 0 \leq B \leq D\}$, does this imply that the elliptic curves that correspond to these lattices are isogenous and therefore one lattice corresponds to an order that it is an over-order (or the same order if the lattices are homothetic) to the order of the other lattice?

Consider two (primitive) lattices $\pi_{i} \in \mathbb{C}$, such that $\pi_{1} = M \pi_{2}$ for $M \in \mathcal{S}_{m}$ with $$\mathcal{S}_{m}:=\Big\{\begin{pmatrix} A & B \\ 0 & D \end{pmatrix} \in \mathrm{GL}_{2}(\mathbb{Z}) : AD = m >0,\;D>0,\;0 \leq B \leq D\Big\}.$$ Does this imply that the elliptic curves that correspond to these lattices are isogenous and therefore one lattice corresponds to an order that it is an over-order (or the same order if the lattices are homothetic) to the order of the other lattice?

If two (primitive) lattices $\pi_{i} \in \mathbb{C}$ are such that $\pi_{1} = M \pi_{2}$ for $M \in \mathcal{S}_{m}:=\{\begin{pmatrix} A & B \\ 0 & D \end{pmatrix} \in GL_{2}(\mathbb{Z}) : AD = m >0, D>0, 0 \leq B \leq D\}$, does this imply that the elliptic curves that correspond to these lattices are isogenous and therefore one lattice corresponds to an order that it is an over-order (or the same order if homothetic) to the order of the other lattice?

If two (primitive) lattices $\pi_{i} \in \mathbb{C}$ are such that $\pi_{1} = M \pi_{2}$ for $M \in \mathcal{S}_{m}:=\{\begin{pmatrix} A & B \\ 0 & D \end{pmatrix} \in GL_{2}(\mathbb{Z}) : AD = m >0, D>0, 0 \leq B \leq D\}$, does this imply that the elliptic curves that correspond to these lattices are isogenous and therefore one lattice corresponds to an order that it is an over-order (or the same order if the lattices are homothetic) to the order of the other lattice?