linear algebra - Order of "one minus automorphism"

Your first condition implies that $r$ is the order of $p$ modulo $q$.

If the order of $1-t$ is $q$, then $p$ divides
$$ prod_{i=1}^{p-1} prod_{j=1}^{p-1} (1 - zeta^i - zeta^j) quad (*)$$
where $zeta$ is a primitive $p$-th root of unity.

If these two conditions are satisfied, we can find $t$ with the required properties.

Proofs:

First, suppose that $t$ obeys your first condition. Let $f$ be the characteristic polynomial of $t$. If $f$ factored as $gh$, and $v$ were in the kernel of $g(t)$, then $mathrm{Span} (t^i v)$ has dimension at most $deg g$, contradicting your hypothesis.

So $f$ is irreducible. In particular, $f$ divides $x^{p^r}-x$ and so $q | p^r-1$. Also, if $f$ divided $x^{p^s} - x$ for some $s<r$, then $f$ has a nontrivial factor of degree $leq s$, contradicting that $f$ is irreducible. So $r$ is the order of $p$ modulo $q$.

Now, suppose that $1-t$ has order $q$. Let $Phi_q(x) = (x^q-1)/(x-1)$. Then $f(x) | Phi_q(x)$ and $f(1-x) | Phi_q(x)$ as well. So $f$ divides $mathrm{GCD}(Phi_q(x), Phi_q(1-x))$. In particular, the resultant of $Phi_q(x)$ and $Phi_q(1-x)$ is zero modulo $p$. The product $(*)$ is precisely that resultant.

Finally, we must reverse the argument. Suppose that $p$ divides $(*)$. So there is an irreducible polynomial $f$ dividing $Phi_q(x)$ and $Phi_q(1-x)$. Let $s = deg f$. Let $V$ be the field $mathbb{F}_{p^s}$ and let $t$ be the action of a root of $f$ on $V$. Then $t$ obeys condition 1. (If not, $mathrm{Span} (t^i v)$ is a $t$-stable subspace of $V$, contradicting that $f$ is irreducible.) By the first paragraph of the proof, this shows that $s$ is the order of $p$ modulo $q$, so $s=r$ and we have constructed the desired $t$.