= Solution
The <Möbius transformations> are bijections, so they preserve which coordinates of a triple are equal. Conversely, part (i) implies transitivity within each equality pattern. For a pattern with fewer than three distinct entries, extend the chosen distinct points to a triple before applying part (i); this works also for $p=2$, when the <projective line> has exactly three points. There are therefore five <orbits of a group action>.
We first count $|GL_2(\mathbb F_p)|$. Its first column can be any nonzero <vector>, giving $p^2-1$ choices. The second must be outside the first column's one-dimensional span, giving $p^2-p$ choices. Hence
$$
|G|=(p^2-1)(p^2-p)=p(p-1)^2(p+1).
$$
For the all-equal pattern take $(\infty,\infty,\infty)$. Its <stabilizer subgroup> consists of upper triangular <matrices>
$$
\begin{pmatrix}a&b\\0&d\end{pmatrix},\qquad a,d\ne0,
$$
of order $p(p-1)^2$. Its <group orbit> has order $p+1$.
The three exactly-two-equal patterns have representatives $(\infty,\infty,0)$, $(\infty,0,\infty)$ and $(0,\infty,\infty)$. Each <stabilizer subgroup> fixes $0$ and $\infty$ individually, so it is the diagonal <subgroup> of order $(p-1)^2$. Each <group orbit> has order $p(p+1)$; their repeated-coordinate positions keep these three orbits distinct.
For the all-distinct pattern use $(0,1,\infty)$. Its <stabilizer subgroup> is the scalar <subgroup> of order $p-1$, and the <group orbit> has order $(p+1)p(p-1)$. Stabilizers of arbitrary triples are conjugates of the displayed representative stabilizers. In summary the <equality-pattern orbits of projective triples> have
$$
\boxed{\begin{array}{c|c|c}
\text{pattern}&\text{orbit size}&\text{stabilizer order}\\\hline
\text{all equal}&p+1&p(p-1)^2\\
\text{each two-equal pattern}&p(p+1)&(p-1)^2\\
\text{all distinct}&p(p+1)(p-1)&p-1
\end{array}}
$$
The sizes add to $(p+1)^3$, and each orbit size times its stabilizer order is $|G|$, as in the <orbit-stabilizer theorem>.
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