= Solution
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 $\operatorname{diag}(a,d)$ with $a,d\ne0$. 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>. Thus
$$
\boxed{\text{There are exactly }p-1\text{ matrices for each distinct target triple.}}
$$
Equivalently, after dividing out <scalar matrices> the <projective general linear group action on the projective line> is <sharply three-transitive on a projective line>.
Back to article page
