Let $\hat{f}: \mathbb{Z}^n \to \mathbb{Z}^k$ be a linear lift of $f$ and let $e_1,\ldots, e_k$ be the elementary divisors of $\text{im}(\hat{f})$ in $\mathbb{Z}^k$.

$|\ker(f)|$ is a power of $m$ iff $\prod_{i=1}^k gcd(m,e_i)$ is a power of $m$ (including $1 = m^0$).

Proof: By the isomorphism theorem, $|\ker(f)|$ is a power of $m$ iff $|\text{im}(f)|$ is a power of $m$. If $A$ is a matrix with integer entries and $\overline{A}$ its reduction modulo $m$ (and similar for vector $x$) then $\overline{A\cdot x} = \overline{A}\cdot \overline{x}$. Hence the image of $f$ equals the reduction of the image of $\hat{f}$ modulo $m$. The image of $\hat{f}$ is $\oplus_{i=1}^k e_i\mathbb{Z}y_i$ for some base $\{y_1,...,y_k\}$. Hence the image of $f$ equals $\oplus_{i=1}^k \bar{e_i}\mathbb{Z}_m\bar{y_i}$. Now the statement follows from $|\bar{e_i}\mathbb{Z}_m|= m/ gcd(m,e_i)$.

Let $\hat{f}: \mathbb{Z}^n \to \mathbb{Z}^k$ be a linear lift of $f$ and let $e_1,\ldots, e_k$ be the elementary divisors of $\text{im}(\hat{f})$ in $\mathbb{Z}^k$.

$|\ker(f)|$ is a power of $m$ iff $\prod_{i=1}^k \gcd(m,e_i)$ is a power of $m$ (including $1 = m^0$).

Proof: By the isomorphism theorem, $|\ker(f)|$ is a power of $m$ iff $|\text{im}(f)|$ is a power of $m$. If $A$ is a matrix with integer entries and $\overline{A}$ its reduction modulo $m$ (and similar for vector $x$) then $\overline{A\cdot x} = \overline{A}\cdot \overline{x}$. Hence the image of $f$ equals the reduction of the image of $\hat{f}$ modulo $m$. The image of $\hat{f}$ is $\oplus_{i=1}^k e_i\mathbb{Z}y_i$ for some base $\{y_1,...,y_k\}$. Hence the image of $f$ equals $\oplus_{i=1}^k \bar{e_i}\mathbb{Z}_m\bar{y_i}$. Now the statement follows from $|\bar{e_i}\mathbb{Z}_m|= m/ \gcd(m,e_i)$.