For the profinite group $\widehat{\mathbb Z} := \varprojlim_n \mathbb Z/n\mathbb Z$ (which ofcourse actually has a ring structure as well), you can show its open subgroups are precisely the groups $n \widehat{\mathbb Z}$ for $n \in \mathbb Z$, and each such subgroups has quotient group $\mathbb Z/n\mathbb Z$. In an informal sense, I'd like to say you can "see" all its open subgroups and their quotients occuring in the defining limit $\varprojlim_n \mathbb Z/n\mathbb Z$.
I am wondering if a similar description of the open normal subgroups and their quotients exist for $\widehat{\mathbb Z}^\times = \varprojlim_n (\mathbb Z/n\mathbb Z)^\times$, or more generally for any profinite group with an explicit descripition as an inverse limit of finite groups, such as for example a profinite completion $\widehat{G} := \varprojlim_N G/N$ of a discrete group (where $N$ ranges over all normal subgroups of finite index).
For $\widehat{\mathbb Z}^\times$, I believe for any $n$ the map $\widehat{\mathbb Z}^\times \to (\mathbb Z/n\mathbb Z)^\times$ is surjective, but I'm not (yet) able to prove this. For this I think it is helpful that the maps $(\mathbb Z/m \mathbb Z)^\times \to (\mathbb Z/n \mathbb Z)^\times$ are surjective if $n \mid m$. Once I know $\widehat{\mathbb{Z}}^\times \to (\mathbb Z/n\mathbb Z)^\times$ is surjective however, I then get an isomorphism $$ \widehat{\mathbb Z}^\times/(1 + n\widehat{\mathbb Z}) \cap \widehat{\mathbb Z}^\times \cong (\mathbb Z/n\mathbb Z)^\times. $$ This is useful since $\widehat{\mathbb Z}$, being profinite, has a neighbourhoodbasis of the identity given by open normal subgroups, which we know are of the form $n \widehat{\mathbb Z}$. Therefore every open subgroup $U$ of $\widehat{\mathbb Z}^\times$ contains $(1 + n\widehat{\mathbb Z}) \cap \widehat{\mathbb Z}^\times$ for some $n$, and therefore $\widehat{\mathbb Z}^\times/U$ must be a quotient of $(\mathbb Z/n\mathbb Z)^\times$.
Now, for any discrete group $G$, form its profinite completion $\widehat{G} = \varprojlim_N G/N$. Then the maps $\widehat{G} \to G/N$ are surjective for every normal subgroup $N$ of $G$, and in this case I know a proof: the composition $G \to \widehat{G} \to G/N$ is the quotient map and therefore surjective, so $\widehat{G} \to G/N$ must also be surjective. Is every open normal subgroup the kernel of such a map $\widehat{G} \to G/N$? Then it would hold that all quotients of $\widehat G$ by an open normal subgroup are isomorphic to $G/N$ for some $N$.