the title says it all. My assumption is, that as I know there is an unique isomorphism between integers and a ring $R$. We have $\phi(n) = n1_{\overline{R}} = (1 + 3i)R$. So, this may mean that $n$ is divisible by $(1 + 3i)$. Here $R$ is the ring of Gauss integers and $I$ is the principal ideal generated by $(1 + 3i)$. But I can't prove it strictly. Maybe I even did this wrong? Or what is your idea about this? How can it be proved?
Write $\ w=1\!+\!3i.\ $ The map $\, \varphi\color{#0a0}{ \ {\rm is\ surjective\ (onto)}}\,$ since $\, \ {\rm mod}\ w\!:\ \,w\bar w=10\equiv 0,\,\ {-}3i\equiv 1\,\overset{\times\,3}\Rightarrow\,i\equiv 3\,\Rightarrow\, a+bi\equiv a+3b\in\Bbb Z$
$\color{#c00}{I = \ker\varphi = 10\,\Bbb Z}\ $ follows by rationalizing a denominator
$\ \ n\in I\iff 1\!+\!3i\mid n\ \, {\rm in}\, \ \Bbb Z[i]\iff \dfrac{n}{1\!+\!3i}\in \Bbb Z[i]\iff \dfrac{n(1\!-\!3i)}{10}\in\Bbb Z[i]\iff \color{#c00}{10\mid n}\qquad$
So applying the First Isomorphism Theorem, $\, \color{#0a0}{R = {\rm Im}\ h} \,\cong\, \Bbb Z/\color{#c00}{\ker h} \,=\, \Bbb Z/\color{#c00}{10\,\Bbb Z}$