The closest I could find was this archived reddit post. There I think they prove that $Z_n\setminus Z_n^{*}$ can be made isomorphic to $Z_m^{*}$.
Question: Let m|n. ($Z_n^{*} \mod m) = Z_m^{*}$ (as sets, not ring isomorphism or any other stronger requirement)?
The "obvious" inclusion is $(Z_n^* \mod m) \subseteq Z_m^{*}$, because if $k$ is coprime with $n$ then it is also coprime with $m$.
Simple true case: the factors of $n$ are all factors of $m$ then $(Z_n^* \mod m) \supseteq Z_m^{*}$, because if a number is coprime with $m$ then it is coprime with $n$ and the result follows from $Z_n^* \supseteq Z_m^{\ast}$.
General case: the problem is when $n/m$ has factors that are not factors of $m$, because then $Z_n^{*} \supseteq Z_m^{*}$ is false (for interesting cases such factors must be smaller than $m$). However, I could not find a counter example; integers $k>m$ in $Z_n^{*}$ always end up generating the integers $k'\in Z_m\setminus Z_n$ after the $\mod m$ operation. Minimal example: $n=6$ and $m=3$, then $Z_6^{*} = \{1,5\}$ and $Z_3^{*}=\{1,2\}$, $2$ is not directly in $Z_n^*$ but it exists as $5$.
Motivation: explaining the invertible cases of modular exponentiation, I recently claimed that finding all the pairs $(x,x^{-1}) \mod \varphi(n)$, the Euler function, and then taking $\mod \lambda(n)$, the Carmichael function, is the same as finding all the pairs $\mod \lambda(n)$ directly. This should be true because it should fall in the simple case, but I claimed it in general, and now I am stuck.
