Why does my transformation sending $0$ to $w$ change in these Möbius transforms? My first question is:

Let $M_{\alpha}, \alpha \in \mathbb{C}$, be the subgroup of $M$ mapping $\alpha$ to itself, that is, the stabilizer of $\alpha$. Given that
$$M_0 = \left \{w = \frac{z}{cz + d}, d \neq 0 \right \}$$
compute the subgroup $M_i, i = \sqrt{-1}$.

Here, $L$ is the transformation sending $0$ to $i$ i,e $L : z \mapsto z + i$
The second one is

Compute the subgroup $M_{\{0, -3, \infty\}} \subset M$ consisting of 6 transformations, preserving the set $\{0, -3, \infty\}$, together with an explicit isomorphism
$$M_{\{0, -3, \infty\}} = S_3$$

Here, my $L: z \mapsto -3z$
In the second one, why does $z \mapsto -3z$ and not $z \mapsto z - 3$ (and therefore why does $z \mapsto z + i$ in the first one)?
 A: In the first case, you want to find $M_i$ given $M_0$. There is a bijection between $M_0$ and $M_i$ given by $T \mapsto L\circ T\circ L^{-1}$ where $L$ is any Möbius transformation such that $L(0) = i$; as you note, $L(z) = z + i$ is one such transformation.
For the second case, you want to find $M_{\{0, -3, \infty\}}$ given $M_{\{0, 1, \infty\}}$. As before, there is a bijection between $M_{\{0, -3, \infty\}}$ given $M_{\{0, 1, \infty\}}$ given by $T \mapsto L\circ T\circ L^{-1}$ where $L$ is any Möbius transformation such that $L(\{0, 1, \infty\}) = \{0, -3, \infty\}$. For $L(z) = z - 3$, we have $L(\{0, 1, \infty\}) = \{-3, -2, \infty\} \neq \{0, -3, \infty\}$. However, if we choose $L(z) = -3z$ then $L(\{0, 1, \infty\}) = \{0, -3, \infty\}$.
In general, for $a, b, c$ distinct, there is a bijection between $M_{\{0, 1, \infty\}}$ and $M_{\{a, b, c\}}$ given by $T \mapsto L\circ T\circ L^{-1}$ where $L$ is any Mobius transformation satisfying $L(\{0, 1, \infty\}) = \{a, b, c\}$; note, for any choice of distinct $a, b, c$, such an $L$ exists.
