(related to Chapter: Rings (Overview))
We look for all endomorphisms (i.e. self-homomorphisms) of the ring $(\mathbb Z,+,\cdot)$ of all integers with addition and multiplication. Let $f:\mathbb Z\to\mathbb Z$ be a ring homomorphism. This means that it has to fulfill the properties $$\begin{array}{rcl} f(x + y)&=&f(x) + f(y),\\ f(x \cdot y)&=&f(x)\cdot f(y)\\ f(1)&=&1. \end{array}$$ Then for the special case $n\ge 1$ we get $$f(n)=f(\underbrace{1+\ldots+1}_{n\text{ times}})=\underbrace{f(1)+\ldots+f(1)}_{n\text{ times}}=\underbrace{1+\ldots+1}_{n\text{ times}}=n.$$ This implies $$f(0)=f(0\cdot 1)=f(0)\cdot f(1)=0\cdot 1=0.$$ Thus $f(n)+f(-n)=0$, or $$f(-n)=-f(n)=-n.$$ It follows that $f:\mathbb Z\to\mathbb Z$ be a ring homomorphism if and only if $f$ is the identity function $f(n)=n$ for all $n\in\mathbb Z.$
The complex conjugation $f:\mathbb C\to\mathbb C$ is a ring homomorphism, since $$\begin{array}{rcl} (z_1+z_2)^*&=&z_1^*+z_2^*\\ (z_1\cdot z_2)^*&=&z_1^*\cdot z_2^*\\ (1)^*&=&1^* \end{array}$$
For any integer $m\ge 2,$ the function $f:\mathbb Z\to \mathbb Z_m$ sith $f(a)=a(m)$ of the commutative ring of integers $(\mathbb Z,+,\cdot)$ to the commutative ring of congruences $\mathbb Z_m$ is a ring homomorphism which is partially shown here. It remains to note that $$1\equiv 1(m)$$ for all $m\ge 2.$
