So in the textbook, we want to map $(x,y,z,1)$ to $\big(\frac{x}{1-\frac{z}{d}}, \frac{y}{1-\frac{z}{d}}, 0, 1\big)$. Since we are working with homogenous coordinates, this is the same thing as mapping $(x,y,z,1)$ to $(x,y,0,1-\frac{z}{d})$. It can be shown the perspective matrix to map this is:
$$P = \begin{bmatrix} 1& 0& 0 & 0 \\ 0& 1& 0 & 0 \\ 0& 0& 0 & 0 \\ 0& 0& -1/d & 1 \\ \end{bmatrix}$$
Now, what I was wondering was if I could just map it from $(x,y,z,1)$ to $\big(\frac{x}{1-\frac{z}{d}}, \frac{y}{1-\frac{z}{d}}, 0, 1\big)$ which gives:
$$P = \begin{bmatrix} \frac{1}{1-\frac{z}{d}}& 0& 0 & 0 \\ 0& \frac{1}{1-\frac{z}{d}}& 0 & 0 \\ 0& 0& 1 & 0 \\ 0& 0& & 1 \\ \end{bmatrix}$$
The final homogeneous vector by applying $P$ to $(x,y,z,1)$ is the same but clearly the matrices $P$ are different. Is there anything fundamentally wrong with doing what I did and have a matrix in terms of the input $z$?
The matrix of your transformation cannot depend on the point, it has to be a constant matrix. There is a one to one correspondence between matrices and transformations (for any given basis).