I would like to see if my answer to this problem is correct or if there are some missing parts.
Automorphism in my textbook means biholomorphism between same domains. i.e $f \in \text{Aut}\left ( \mathbb{C}\setminus \{1\} \right) = \{f:\mathbb{C}\setminus \{1\} \rightarrow \mathbb{C}\setminus \{1\} : f \text{ is biholomorphism} \}$.
First, let $f\in \text{Aut}\left ( \mathbb{C}\setminus \{1\} \right )$ be arbitrarily chosen. Letting $g(z)=f(z+1)-1$, we can see that $g\in \text{Aut}\left ( \mathbb{C}\setminus \{0\} \right )$.
By the Theorem which tells us "$g\in \text{Aut}\left ( \mathbb{C}\setminus \{0\} \right ) \Leftrightarrow \exists a\in \mathbb{C}\setminus \{0\}$ s.t either $g(z)=az$ or $g(z)=\frac{1}{az}$," we get $f(z+1)=g(z)+1$ so that $f(z+1)=az+1$ or $f(z+1)=\frac{1}{az}+1$.
This implies that either $f(z)=az-a+1$ or $f(z)=\frac{1}{a(z-1)}+1$.
Hence, every element in $\text{Aut}\left ( \mathbb{C}\setminus \{1\} \right )$ has to be in the form of $az-a+1$ or $\frac{1}{a(z-1)}+1$ with $a \neq 0$.
Any help finding flaws in my solution would be really appreciated.