The only function that forms a homomorphism from $\mathbb Z_n$ to $\mathbb Z_n$

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Define $M_a : Z_n → Z_n$ by $M_a([x]) = [ax]$

Prove that

(a) $M_a$ : $(Z_n, +) → (Z_n, +)$ is a homomorphism.

(b) Let φ : $(Z_n, +) → (Z_n, +)$ be a homomorphism. Prove that φ = $M_a$ for some a ∈ Z.

(c) For two homomorphism φ, ψ : $(Z_n, +) → (Z_n, +)$, prove that if φ = $M_a and ψ = M_b$, then for their composition φ ◦ ψ = $M_{ab}$.

My attempt: (a) is just using the definition of homomorphism and knowing that $[ax+ay]_n=[ax]_n +[ay]_n$

(c) involved simply using definitions of these functions as well.

I do not know how to prove (b): Using the fact that the function has been given to be a homomorphism, what I do is:

$ψ(0 + a) = ψ(0) + ψ (a)$

This means $ψ (a) = [0]_n$ But I do not see how this will help me conclude the function must be of the form $M_a$ Could someone please help or provide an alternative?

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Let $a\in\mathbb Z$ be such that $\varphi(1)=[a]$. Then $\varphi(1)=M_a(1)$. Since $\mathbb{Z}_n=\langle1\rangle$, you deduce from this that $\varphi=M_a$.