If there is an injective ring homomorphism of $\Bbb{Z_5}$ to $\Bbb{Z}_n$ then $n$ is .... 
Question. If there is an injective ring homomorphism of $\Bbb{Z_5}$ to $\Bbb{Z}_n$ then $n$ is ....

I want to find the possible values of $n$ in this situation. 
My attempt. $\Bbb{Z}_5$ is a field and if there exists such injective ring homomorphism then $\Bbb{Z_n}$ should contains a copy of $\Bbb{Z}_5$. I cannot proceed further....Any help is appreciated. Thank you.
 A: If you allow nonunital ring homomorphisms, then there is an injective map $\phi : \Bbb Z_5\to\Bbb Z_n$ if and only if $n = 5m,$ where $5$ and $m$ are relatively prime.
Write $n = 5^r m,$ where $r\geq 1$ and $(5,m) = 1.$ In order to obtain a (perhaps nonunital) ring homomorphism $\Bbb Z_5\to\Bbb Z_n,$ we need to map $1$ to an idempotent $f\in\Bbb Z_n$ such that $5f \equiv 0\pmod{n}.$ $5f\equiv 0\pmod{n}$ implies that $f\equiv 0\pmod{5^{r-1}m}.$ It follows that $5^{r-1}\mid f$ and that if $p$ is a prime dividing $m$ with multiplicity $e_p,$ that $p^{e_p}\mid f.$
In order for $f$ to be an idempotent in $\Bbb Z_n,$ you must have $f\equiv 0$ or $f\equiv 1$ mod $5^r$ and mod $p^{e_p}$ for all $p$ dividing $m$ with multiplicity $e_p$ (see here). If $r > 1,$ it follows that $f\equiv 0\pmod{5^r},$ as $f = 5^{r-1}m$ can never be equivalent to $1$ mod $5^r.$ But then $f\equiv 0\pmod{5^r}$ and $f\equiv 0\pmod{p^{e_p}}$ for all $p$ dividing $m,$ and so $f$ is $0.$ (In this case, the corresponding map will send every element of $\Bbb Z_5$ to $0$ - very non-injective.)
Thus, we have deduced that it is necessary that $n$ be of the form $5m,$ where $(5,m) = 1.$ It is also sufficient: in this case, we may use the Chinese remainder theorem to find an $f$ such that $f\equiv 1\pmod{5}$ and $f\equiv 0\pmod{p^{e_p}}$ for any $p$ dividing $m$ with multiplicity $e_p.$ This is a nonzero idempotent of $\Bbb Z_n,$ and by construction, we will have $5f\equiv 0\pmod{n}.$
It remains to be justified that the map $1\mapsto f$ where $f$ is of the form described in the previous paragraph is injective. Indeed, suppose $rf\equiv 0\pmod{n}.$ Then as $f\equiv 0\pmod{m},$ it follows that $f = mk$ for some $k.$ Thus, $rf\equiv rmk\equiv 0\pmod{5m}\implies rk\equiv 0\pmod{5}.$ As $f\equiv mk\equiv 1\pmod{5},$ it follows that $k$ is a unit mod $5,$ and hence that $r\equiv 0\pmod{5}.$ So if $rf\equiv 0\pmod{n},$ we have $5\mid r,$ and by construction, $5f\equiv 0\pmod{n}.$ Therefore, we deduce that the additive order of $f$ is $5,$ so that the map $1\mapsto f$ is injective.
As an example of such a nonunital ring homomorphism, consider the map
\begin{align*}
\phi : \Bbb Z_5&\to\Bbb Z_{10}\\
1&\mapsto 6\\
2&\mapsto 2\\
3&\mapsto 8\\
4&\mapsto 4\\
5&\mapsto 0.
\end{align*}
