So I am being asked how many $\mathbb{Z}[x]$ module structures does the abelian group $\mathbb{Z}/5\mathbb{Z}$ have. Thinking about it, we really only need to define the action of $x$ on $[1]$ because if we do so, then we know what the action of $\sum_{i=0}^n a_ix^i$ would be on $[m]$ by:$(\sum_{i=0}^na_ix^i)\cdot [m]=(\sum_{i=0}^na_ix^i)\cdot[1+...+1]=\sum_{i=0}^na_i(x^i\cdot [1+...+1])$and if we have that $x\cdot [1]=[t]$, then we would have that $x^2\cdot [1]=x\cdot[t]=x\cdot[1+...+1]=x\cdot[1]+...+x\cdot[1]=[t]+...+[t]=[t^2]$ and in general, $x^i\cdot [1]=[t^i]$, note that we also have that the action of $\mathbb{Z}$ on $\mathbb{Z}/5\mathbb{Z}$ is fixed by $a\cdot[m]=[am]$. Hence, if we know $x\cdot [1]$ we would know the entire action of our ring on our module. Looking at it, I think that we can let $x\cdot [1]$ be equal to anything and I dont see what problems can be caused, so my answer would be that there are $5$ possible structures.
I see the similarity of having $f(x)\in \mathbb{Z}[x]$ and then taking $k\in \mathbb{Z}$ and doing $[f(k)]$ modulo $5$, but I dont know if there is any meaning behind it or what.