How many ways can $\mathbb{Z}/5\mathbb{Z}$ be given the structure of a module over Gaussian integers? As the title says, the question is: In how many ways can the additive group $\Bbb{Z}/5\Bbb{Z}$ be given the structure of a module over Gauss integers?
My attempt:
Any $\Bbb{Z}[i]$-module M is an abelian group with homomorphism $\psi: M \rightarrow M$ such that $\psi^2 = -I_M$. 
Let $F = \Bbb{Z}/5\Bbb{Z}$. $F$ is a cyclic group generated by $5$, so the homomorphism $\psi: F \rightarrow F$ is determined by $\psi(5)$.
If $\psi^2 = -I_M$, then $a^2 = -1$ in $F$. So find $a \in \Bbb{Z}$ such that $a^2 \equiv -1$ (mod $5$). And the number of $a$'s that satisfy this condition is equal to the number of ways we can give $F$ the structure of a module over Gaussian integers.
I feel like this answer makes sense but I'm not sure if I did this correctly since my answer gives the conditions needed but not the definitive number. Any suggestions/tips would be greatly appreciated.
 A: An $R$-module structure on the abelian group $G$ is a ring homomorphism $R\to\operatorname{End}(G)$. Since $\operatorname{End}(\mathbb{Z}/5\mathbb{Z})\cong\mathbb{Z}/5\mathbb{Z}$ as rings, what you want is the number of ring homomorphisms $\mathbb{Z}[i]\to\mathbb{Z}/5\mathbb{Z}$.
Such a homomorphism must send $1$ to $[1]$ (as we want unital modules, don't we?) and $i$ to some element $u$ of $\mathbb{Z}/5\mathbb{Z}$ such that $u^2=-[1]=[4]$. You have two choices: $u=[2]$ or $u=-[2]=[3]$.
Both define a ring homomorphism, as it can be checked explicitly or, better, by first defining $f\colon\mathbb{Z}[X]\to\mathbb{Z}/5\mathbb{Z}$ with $f(X)=[2]$ (or $[3]$) and checking that the kernel of this homomorphism contains the ideal generated by $X^2+1$.

Alternatively, you can find the endomorphisms $\psi\colon\mathbb{Z}/5\mathbb{Z}\to\mathbb{Z}/5\mathbb{Z}$ such that $\psi^2=-I_{\mathbb{Z}/5\mathbb{Z}}$. Such an endomorphism is determined by the image of $[1]$, which must be either $[2]$ or $[3]$.

Notes.


*

*With $[a]$ I denote the residue class of $a$ in $\mathbb{Z}/5\mathbb{Z}$. 

*You're wrong in stating that $\mathbb{Z}/5\mathbb{Z}$ is generated by $5$ (or $[5]$). Instead it is generated by $[1]$ (or any other element different from $[5]=[0]$).
