One set of condition is that the matrix $A$ is nilpotent (which can be read off from the characteristic polynomial $p_A(t)$ being $t^n$) and that its cokernel is isomorphic to $\mathbb Z$ (or equivalently its elementary divisors are $n$ $1$'s and one $0$). Indeed considering (through $A$) $\mathbb Z^n$ as a module over $\mathbb Z[t]/(t^n)$ and letting $e\in\mathbb Z^n$ be a lift of a generator of $\mathbb Z^n/A\mathbb Z^n$ we get by Nakayama's lemma that it is a generator of $\mathbb Z^n$ as an $\mathbb Z[t]/(t^n)$-module giving us a surjective $\mathbb Z[t]/(t^n)$-module-map $\mathbb Z[t]/(t^n)\rightarrow\mathbb Z^n$. It is also injective as that can be checked over $\mathbb Q$ in which case it is clear as $A$ as $\mathbb Q$-matrix has a single Jordan block.
As for the more general question a Jordan block of the assumed type has cokernel isomorphic to $\mathbb Z\bigoplus \mathbb Z/b_1\bigoplus\cdots\bigoplus\mathbb Z/b_{n-1}$. I haven't really thought about it carefully but I doubt that that is enoughthe only invariant.
One set of condition is that the matrix $A$ is nilpotent (which can be read off from the characteristic polynomial $p_A(t)$ being $t^n$) and that its cokernel is isomorphic to $\mathbb Z$ (or equivalently its elementary divisors are $n$ $1$'s and one $0$). Indeed considering (through $A$) $\mathbb Z^n$ as a module over $\mathbb Z[t]/(t^n)$ and letting $e\in\mathbb Z^n$ be a lift of a generator of $\mathbb Z^n/A\mathbb Z^n$ we get by Nakayama's lemma that it is a generator of $\mathbb Z^n$ as an $\mathbb Z[t]/(t^n)$-module giving us a surjective $\mathbb Z[t]/(t^n)$-module-map $\mathbb Z[t]/(t^n)\rightarrow\mathbb Z^n$. It is also injective as that can be checked over $\mathbb Q$ in which case it is clear as $A$ as $\mathbb Q$-matrix has a single Jordan block.
As for the more general question a Jordan block of the assumed type has cokernel isomorphic to $\mathbb Z\bigoplus \mathbb Z/b_1\bigoplus\cdots\bigoplus\mathbb Z/b_{n-1}$. I haven't really thought about it carefully but I doubt that that is enough.