I want to convince myself of the following proposition:
Prop. For distinct $p_{1},p_{2},p_{3}\in\mathbb{P}^{1}_{\mathbb{C}}$ and distinct $p_{1}',p_{2}',p_{3}'\in\mathbb{P}^{1}_{\mathbb{C}}$ there is a unique automorphism of $\mathbb{P}^{1}_{\mathbb{C}}$ taking $p_{i}$ to $p_{i}'$.
I have looked at several MathStackExchange posts as well as this set of notes. From what I gather so far, we can understand automorphisms of $\mathbb{P}^{1}_{\mathbb{C}}$ as maps $[x:y]\mapsto[ax+by:cx+dy]$ for some $a,b,c,d\in\mathbb{C}$ and subject to the constraint $ac-bd\neq0$. It is routine to verify that this map respects scaling by $\lambda\in\mathbb{C}^{*}$ as follows $$[x:y]=[\lambda x:\lambda y]\mapsto [\lambda ax+\lambda by:\lambda cx+\lambda cy]=\lambda[ax+by:cx+dy]=[ax+by:cx+dy]$$ What I don't understand is how this set of matrices is "three dimensional" as stated in the notes. Moreover, while I understand intuitively how the "three dimensionality" would imply a unique automorphism of $\mathbb{P}^{1}_{\mathbb{C}}$ (since each $p_{i}$ mapping to $p_{i}'$ would introduce a constraint to one of the dimensions) but am not exactly sure how to construct the appropriate matrix giving said transformation.
To get an automorphism, we need both $\Gamma = \{p_1, p_2, p_3\}$ and $\Gamma'= \{p_1', p_2', p_3'\}$ to be in general position, which is to say that any two elements of $\Gamma$ or any two elements of $\Gamma'$ span the underling $k^2$ when we write $\mathbb{P}^1 = \mathbb{P}(k^2)$. Now, any three points of $\mathbb{P}^1$ in general position form what's called a projective basis for $\mathbb{P}^1$. To see this, let $v_1, v_2$ be representatives in the affine pre-quotiented $k^2$ for $p_1$ and $p_2$, respectively. Since $p_1, p_2$ are in general position they span the underlying $k^2$ so we can write $$ v_3 = \sum_{i=1}^2 \lambda_i v_i $$ for a fixed choice of representative $v_3$ for $p_3$. What this means is we can now re-scale $v_1, v_2$ to \begin{align*} \tilde{v_1} &= \frac{1}{\lambda_1} v_1 & \text{ and }\\ \tilde{v_2} &= \frac{1}{\lambda_2} v_2 \end{align*} so that $v_3 = \tilde{v}_1 + \tilde{v}_2$. That is, without loss of generality, $p_3 = p_1 + p_2$. Now, suppose we want an automorphism of $\mathbb{P}^1$ that maps $p_1 \mapsto [1:0], p_2 \mapsto [0:1]$ and $p_3 \mapsto [1:1]$. Then just the re-scaled $v_1$ and $v_2$ give
\begin{align*} A = \left[\begin{matrix} \uparrow & & \uparrow \\ \tilde{v_1} & & \tilde{v_2} \\ \downarrow & & \downarrow \end{matrix}\right] \end{align*} as an automorphism of the underlying $k^2$ which maps $e_i \mapsto \tilde{v}_i$. By construction, $A(v_3) = A(\tilde{v_1}+\tilde{v_2})$.
Correspondingly, $A^{-1}$ mapsto $\tilde{v}_i \mapsto e_i$ for $1\leq i \leq 2$, and $(1,1)\mapsto v_3$. Now, why does this give an equivalence class $[A] \in PGL_2(k) = GL_2(k)/(\lambda\cdot \text{Id}) $? The entire story of re-scaling $v_1, v_2$ for a given choice of $v_3$ as representatives of equivalence classes in the affine pre-quotiented $k^2$ is invariant under $\lambda\in k^*$. Thus there's a unique automorphism $[A^{-1}]$ of $\mathbb{P}^1$ carrying $p_1,p_2,p_3$ to $[1:0], [0:1]$, and $[1:1]$ as a statement about the uniqueness of linear automorphism of the underlying $k^2$ (modulo re-scaling by $\lambda\in k^*$). Similarly, there's a unique automorphism $[A'^{-1}]$ of $\mathbb{P}^1$ carrying $p_1', p_2', p_3'$ to $[1:0], [0:1]$, and $[1:1]$. The statement follows.