I wanted some feedback on the validity of the following proof and advices on its formulation.
For each $n \in \mathbb{N}$, let $\mu_n := \{z \in \mathbb{C} : z^n = 1\}$. Find all homomorphisms $\varphi: \mu_{12} \to \mu_6$.
If $\varphi:\mu_{12} \to \mu_6$ is a homomorphism then $\text{Im }\varphi \leq \mu_6$. By Lagrange's Theorem, the only possible subgroups of $\mu_6$ have orders $1,2,3$ or $6$. Thus way, for each $k \in \{2,4,6,12\}$ define $\varphi_k:\mu_{12} \to \mu_6$, $$\varphi_k (\zeta) = \zeta^k.$$
Observe that $\varphi_k(\zeta)^6 = (\zeta^k)^6 = (\zeta^{k/2})^{12} = 1 \in \mu_6$, showing that $\varphi_k$ is well-defined. To show $\varphi_k$ is indeed a homomorphism, let $\zeta,\xi \in \mu_{12}$, and therefore $$\varphi_k(\zeta\xi) = (\zeta\xi)^6 = \zeta^6\xi^6 = \varphi_k(\zeta)\varphi_k(\xi).$$
Notice that $\text{ord} (\text{Im } \varphi_k) = 12/k$, which are the orders of all the possible subgroups of $\mu_6$ and therefore $\varphi_k$, $k \in \{1,2,6,12\}$, are all the homomorphisms from $\mu_{12}$ to $\mu_6$.
The proof is not correct. In particular, $$\varphi_k(\zeta)=\zeta^k$$ is always a homomorphism $\mu_{12}\rightarrow \mu_{12}$, and this essentially follows from your argument. What you miss is that $\varphi_k$ is a homomorphism from $\mu_{12}\rightarrow\mu_6$ for all even $k$ - so you missed $k=8$ or $k=10$ in particular.
While you are correct that the image of a homomorphism is a subgroup of $\mu_6$ and the only possible orders of a subgroup are $\{1,2,3,6\}$, you make an error in assuming that the order of the image of a homomorphism characterizes the homomorphism. In particular $\varphi_2$ and $\varphi_{10}$ both have the same image, but are distinct homomorphisms.