A projective tranformation on $\mathbb{P}^{1 }$ is a map $P(f): P(V) \to P(V)$ induced by a bijective linear map $f: V-\{0\} \to V-\{0\}$, where $V$ is a two dimensional vector space.
Any linear map $f: V \to V$ is entirely described by where it maps two vectors, as long as those two vectors do not span the same line, i.e are linearly independent.
Suppose I lift two distinct points $p_1$, and $p_2$ in $\mathbb{P}(V)$ to representatives of their classes in $V$. These two points will necessarily be linearly independent in $V$. Suppose I do the same for $q_1$ and $q_2$ in $\mathbb{P}(V)$. Then their exists a unique linear automorphism $A:V \to V$ taking $[p_1] \to [q_1]$ and $[p_2] \to [q_2]$.
Now, I have it in my mind, that this induced a unique projective transformation $P(A): \mathbb{P}(V) \to \mathbb{P}(V)$ up to multiplcation by a scalar diagonal matrix, i.e induces a unique element of $PGL$.
However, this is not true as it is common knowledge that we must determine the image of three points in $P(V)$ and not just two.
What am I missing?
Once you've chosen representative vectors for the four projective points $p_1,p_2,q_1,q_2$, there will be just one linear transformation sending the representative vectors of the $p_i$'s to those of the $q_i$'s. But there will be a different linear transformation sending your chosen representatives of $p_1$ and $p_2$ to $3$ times the chosen representative of $q_1$ and $7$ times the chosen representative of $q_2$. The associated projective transformation will send $p_1$ and $p_2$ to $q_1$ and $q_2$, since the factors $3$ and $7$ don't affect the projective points. But that second linear transformation will not have the same effect as the first on other projective points.