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The general linear group over a finite field acting diagonally on projective triples has five orbits of a group action: all equal, three different positions for the repeated coordinate when exactly two are equal, and all distinct. The corresponding representative stabilizer subgroups are upper triangular, diagonal, and scalar matrices. Their orders are $p(p-1{)}^2$, $(p-1{)}^2$, and $p-1$, respectively; the orbit sizes are $p+1$, $p(p+1)$, and $p(p+1)(p-1)$. Transitivity follows from the action being sharply three-transitive on a projective line.

Identify the projective line with one-dimensional subspaces of ${\mathbb{F}}_p^2$: use $v_t=(t,1{)}^T$ for finite $t$, and $v_{\infty }=(1,0{)}^T$. This convention turns the matrix action into the stipulated Möbius transformation.

Since $x$ and $z$ are distinct projective points, $v_z,v_x$ are a basis. Write $v_y=A{v}_z+B{v}_x$. Both $A$ and $B$ are nonzero, because $y$ is different from $x,z$. The matrix with columns $A{v}_z$ and $B{v}_x$ is invertible and sends the lines of $(1,0{)}^T,(0,1{)}^T,(1,1{)}^T$ to those of $z,x,y$, respectively. This constructs the requested map without separate exceptional formulas at infinity.

A matrix fixing $0$ and $\infty$ is diagonal, say $\mathrm{diag}(a,d)$ with $a,d\ne 0$. To fix $1$ it must also satisfy $a=d$, so the stabilizer subgroup of $(0,1,\infty )$ is precisely $\{\lambda I:\lambda \in {\mathbb{F}}_p^{\times }\}$. Any two solutions differ on the right by one of these matrices. ThusEquivalently, after dividing out scalar matrices the projective general linear group action on the projective line is sharply three-transitive on a projective line.

A scalar matrix is a scalar multiple of the identity matrix, equivalently a diagonal matrix with all diagonal entries equal. It commutes with every square matrix of the same size. In $G{L}_n(F)$, the nonzero scalar matrices form the central subgroup removed to define the projective linear group.