= Solution
Every element of $GL_2(\mathbb F_q)$ is conjugate to exactly one representative of one of these types, with the indicated parameter identifications:
* $aI$, where $a\in\mathbb F_q^\times$;
* $\begin{pmatrix}a&1\\0&a\end{pmatrix}$, where $a\in\mathbb F_q^\times$;
* $\operatorname{diag}(a,b)$, where $a,b\in\mathbb F_q^\times$, $a\neq b$, and $(a,b)$ is unordered;
* $\begin{pmatrix}0&-N(\lambda)\\1&\operatorname{Tr}(\lambda)\end{pmatrix}$, where $\lambda\in\mathbb F_{q^2}\setminus\mathbb F_q$ is identified with $\lambda^q$.
The projective line $\mathbb P^1(\mathbb F_q)$ has $q+1$ points. A line is fixed exactly when it is an eigenline over $\mathbb F_q$, so the permutation character $\pi$ takes respectively the values
$$
q+1,\qquad1,\qquad2,\qquad0.
$$
The action is two-transitive. Part (a) therefore gives
$$
\pi=1_G+\mathrm{St},
$$
where the nontrivial constituent $\mathrm{St}$ is irreducible of degree $q$, the <Steinberg representation of GL2 over a finite field>. Its values on the four types are
$$
\boxed{q,\qquad0,\qquad1,\qquad-1.}
$$
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