 Solution

The PDF has $q$-power exponents; the polynomial is a linearized polynomial over a finite field, not the degree-$d$ polynomial produced by the converted TeX. Let $V$ be its root set in an algebraic closure. Because $(x+y{)}^{q^i}=x^{q^i}+y^{q^i}$ and $c^{q^i}=c$ for $c\in {\mathbb{F}}_q$,Thus $V$ is an ${\mathbb{F}}_q$-vector space. Its derivative is $a_0\ne 0$, and its degree is $q^d$, so it has exactly $q^d$ distinct roots. A finite vector space of that size has dimension $d$.

Every splitting-field automorphism fixes ${\mathbb{F}}_q$ and carries roots to roots. Hence it acts linearly on $V$. The action is faithful because the splitting field is generated by the roots. Choosing a basis identifies this injection withDifferent bases conjugate the matrix subgroup. The linear action itself is canonical, while a particular matrix realization is basis-dependent.