It is enough to show that

if $M\subseteq \mathbb Z^3$ be a subgroup such that $\mathbb Z^3/M$ is a cyclic group of order $k$, then there exists $g\in\mathrm{SL}(3,\mathbb Z)$ such that $g(M)=\langle (1,0,0),(0,1,0),(0,0,k)\rangle$.

Let $M\subseteq \mathbb Z^3$ be a subgroup such that $\mathbb Z^3/M$ is a cyclic group of order $k$. Then $M$ is free of rank $3$, and there exists $A\in M(3,\mathbb Z)$ such that $M=A\cdot\mathbb Z^3$. Using the Smith normal form, we know that there exists $3\times 3$ matrices $P$ and $Q$, invertible over $\mathbb Z$, such that $PAQ=D$ with $D=\left(\begin{smallmatrix}a\\\&b\\\&&c\end{smallmatrix}\right)$ and $a\mid b\mid c$. Then $PM=PAQ\mathbb Z^3=D\mathbb Z^3$.

It follows that $P\in\mathrm{SL}(3,\mathbb Z)$ is such that $PM$ is generated by $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$ with $a\mid b\mid c$. Since $\mathbb Z^3/g(M)$ is cyclic of order $k$, we must have $a=b=1$ and $c=k$. This tells us that the claim above is true.

It is enough to show that

if $M\subseteq \mathbb Z^3$ be a subgroup such that $\mathbb Z^3/M$ is a cyclic group of order $k$, then there exists $g\in\mathrm{SL}(3,\mathbb Z)$ such that $g(M)=\langle (1,0,0),(0,1,0),(0,0,k)\rangle$.

Let $M\subseteq \mathbb Z^3$ be a subgroup such that $\mathbb Z^3/M$ is a cyclic group of order $k$. Then $M$ is free of rank $3$, and there exists $A\in M(3,\mathbb Z)$ such that $M=A\cdot\mathbb Z^3$. Using the Smith normal form, we know that there exists $3\times 3$ matrices $P$ and $Q$, invertible over $\mathbb Z$, such that $PAQ=D$ with $D=\left(\begin{smallmatrix}a\\\&b\\\&&c\end{smallmatrix}\right)$ and $a\mid b\mid c$. Then $PM=PAQ\mathbb Z^3=D\mathbb Z^3$.

It follows that $P\in\mathrm{SL}(3,\mathbb Z)$ is such that $PM$ is generated by $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$ with $a\mid b\mid c$. Since $\mathbb Z^3/g(M)$ is cyclic of order $k$, we must have $a=b=1$ and $c=k$. This tells us that the claim above is true.

(I've done everything at the level of generality which your problem needs, and I'll leave the fun of finding the correct general statement for you...)

Let $M$ be a subgroup of $\mathbb Z^3$ such that $\mathbb Z^3/M$ is cyclic of order $k$. Then $M$ is free of rank $3$, so there is a matrix $A\in M_3(\mathbb Z)$ of non-zero determinant such that $M=A\cdot\mathbb Z^3$. Then there exists $3\times 3$ matrices $P$ and $Q$, invertible over $\mathbb Z$, such that $PAQ$ is a diagonal matrix with entries $a$, $b$, $c$ such that $a\mid b\mid c$, and the three numbers $a$, $b$, $c$ depend only on $M$.

It follows that there is an element $g\in\mathrm{SL}(3,\mathbb Z)$ such that $g(M)$ is generated by $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$. Now, since $\mathbb Z^3/g(M)$ is cyclic of order $k$, we must have $a=b=1$, $c=k$.

It is enough to show that

if $M\subseteq \mathbb Z^3$ be a subgroup such that $\mathbb Z^3/M$ is a cyclic group of order $k$, then there exists $g\in\mathrm{SL}(3,\mathbb Z)$ such that $g(M)=\langle (1,0,0),(0,1,0),(0,0,k)\rangle$.

Let $M\subseteq \mathbb Z^3$ be a subgroup such that $\mathbb Z^3/M$ is a cyclic group of order $k$. Then $M$ is free of rank $3$, and there exists $A\in M(3,\mathbb Z)$ such that $M=A\cdot\mathbb Z^3$. Using the Smith normal form, we know that there exists $3\times 3$ matrices $P$ and $Q$, invertible over $\mathbb Z$, such that $PAQ=D$ with $D=\left(\begin{smallmatrix}a\\\&b\\\&&c\end{smallmatrix}\right)$ and $a\mid b\mid c$. Then $PM=PAQ\mathbb Z^3=D\mathbb Z^3$.

It follows that $P\in\mathrm{SL}(3,\mathbb Z)$ is such that $PM$ is generated by $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$ with $a\mid b\mid c$. Since $\mathbb Z^3/g(M)$ is cyclic of order $k$, we must have $a=b=1$ and $c=k$. This tells us that the claim above is true.