For $n\in \mathbb{Z}$, define the map $f_n:S^1\to S^1$ as $f_n(z)= z^n$, where the unit circle $S^1$ is observed as the subspace $\{z\in\mathbb{C}|\ |z|=1\}$.
How would one compute the induced group homomorphism $(f_n)_*\pi_1(S^1,1)\to \pi_1(S^1,1)$ with the identification of $\pi_1(S^1,1)$ with $\mathbb{Z}$?
On
You would start out with an explicit description of the isomorphism between $\mathbb Z$ and $\pi_1(S^1, 1)$, i.e. you would write down a loop $\alpha$ representing the class that corresponds to $1\in\mathbb Z$. Then you would see what happens to that class under $(f_n)_\ast$, i.e. you would look at $f_n\!\circ\!\alpha$. You would try to express this loop in terms of $\alpha$. You might try to see if drawing images for $n=-1,0,1,2,3$ helps.
$f : S^1 \to S^1$ given by $z \mapsto z^n$ is a covering map for $n \neq 0$, which implies that the induced map $\pi_1 f : \pi_1(S^1) \to \pi_1(S^1)$ is injective. In fact, $\pi_1f$ sends the generator of $\pi_1(S^1)$ to the element in $\pi_1(S^1)$ given by the loop $\gamma : S^1 \to S^1$ defined as $z \mapsto z^n$ which is $n$-times the generator of $\pi_1(S^1)$ of the base copy.
Thus, the map $\pi_1 f$ is the homomorphism $\Bbb Z \stackrel{\times n}{\longrightarrow} \Bbb Z$ given by sending the generator $a$ to $na$.
For $n = 0$, the map $f : S^1 \to S^1$ is the constant map, and thus the induced homomorphism $\pi_1 f$ is the zero homomorphism $\pi_1 f : \Bbb Z \to 0$.