= 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 $c\in\mathbb F_q$,
$$
f(x+y)=f(x)+f(y),\qquad f(cx)=c f(x).
$$
Thus $V$ is an $\mathbb F_q$-vector space. Its derivative is $a_0\ne0$, 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 with
$$
\boxed{\operatorname{Gal}(f,\mathbb F_q(t))\hookrightarrow GL(d,q).}
$$
Different bases conjugate the matrix subgroup. The linear action itself is canonical, while a particular matrix realization is basis-dependent.
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