Let $G=\{\begin{pmatrix} a&b \\ c&d \end{pmatrix}, a, b, c, d \in \mathbb{F_p}, ad-bc \neq 0 \}$.
Let $G$ act on $X=\{ \begin{pmatrix} x \\ y \end{pmatrix}, x, y \in \mathbb{F_p}\}$ by $\begin{pmatrix} a&b \\ c&d \end{pmatrix}\cdot\begin{pmatrix} x \\ y \end{pmatrix}=\begin{pmatrix} ax+by \\ cx+dy \end{pmatrix}$.
Let $g$ an element of order $p$ in $G$. By using orbit-stabilizer show that there exist $x, y \in \mathbb{F_p}$, not both $0, $ s.t. $g \cdot \begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x \\ y \end{pmatrix}$ and deduce that $g$ is conjugate to $\begin{pmatrix} 1&1 \\ 0&1 \end{pmatrix}$.
What confuses me is that what the question is asking to prove is that for a fixed element $g$ of order $p$, there is some element $x\in X$ stabilised by it and I can't see how to apply orbit-stabilizer to show that. Maybe it has to do with the fact that the action is transitive?
The group $\langle g\rangle$ of order $p$ acts on $X$, and this action partitions $X$ into orbits of lengths dividing $p$. Because $|X|=p^2$ and the orbit of $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ has length $1$, there must be at least $p-1$ other orbits of length $1$. This means there is a nonzero point in $X$ that is stabilized by $g$.