Behind this bushy statement stands the following straightforward question. Given a prime $p$ such that $p-1 = 2q$ with $q$ prime, find positive integers $k\geq q-1 = \frac{p-1}{2}-1$ and $m$ such that $(1^{m^i}, 2^{m^i},\ldots,(p-1)^{m^i})$ are distinctly ordered complete sets of residues modulo $p$ for all $0\leq i \leq k-1$. This forces $\gcd(m,p-1)=1$. Moreover, since $\mathbb{Z}_{p-1}^* = \mathbb{Z}_{2q}^*$ is known to be cyclic, of order $\varphi(2q) = q-1$, it follows that $k\leq q-1$, thus we will prove the thesis for $k=q-1$. It is now enough to take $m$ a generator of $\mathbb{Z}_{2q}^*$, so that $m^i \not \equiv 1 \pmod{p-1}$ for all $1\leq i < q-1$ (but $m^{q-1} \equiv 1 \pmod{p-1}$). For $g$ a generator of $\mathbb{Z}_{p}^*$, this makes that $g^{m^i}$ are distinct modulo $p$ for all $0\leq i \leq q-1$, thus the ordered complete sets $(1^{m^i}, 2^{m^i},\ldots,g^{m^i},\ldots,(p-1)^{m^i})$ of residues modulo $p$ are distinct.