The first paragraph here presumes that for every finite field, there is some element whose powers generate all non-0 field elements.
For the field with p^{q} elements there are p^{q}−1 such non-0 elements to cycle thru and this number is seldom prime.
How are we to know that there is such a generator?
The non-0 elements form an abelian group under multiplication and not all such have generators, witness the Vierergruppe.
Luckily GF(5) has the cyclic group as its multiplicative group and cyclic groups have generators.
Does our luck hold out?
Any group with an element that generates all others is (isomorphic to) the cyclic group of that order.
The implicit claim seems to be that the multiplicative groups of all finite fields are isomorphic to the cyclic group.
There is another source of automorphisms of GF(p^{q}).
If we represent GF(p^{q}) computationally by q-tuples of integers mod p, then we must first choose a primitive polynomial of degree q over GF(p).
Of all polynomials of degree q, about 1/q are primitive.
Let x and y be two such polynomials.