Let $p$ be an odd prime, $g$ a primitive root of $p$ and $A=\{1, 2, \ldots, p-1 \}$.

Obviously, $\sigma_g(p)=\begin{pmatrix} 1 & 2 & \ldots & {p-1} \\ g^1\pmod{p} & g^2\pmod{p} & \ldots & g^{p-1}\pmod{p} \end{pmatrix}$ is a permutation of $A$.

I observed that "almost always" $\sigma$ is a product of cycles whose length is strictly less than $p-1$.

If $p=3$ and $g=2$, then $\sigma_2(3)=\begin{pmatrix}
1 & 2 \\
2 & 1
\end{pmatrix}=(1\quad2)$ is a cycle of length $3-1=2$.

If $p=5$ and $g=3$, then $\sigma_3(5)=\begin{pmatrix}
1 & 2 & 3 & 4 \\
3 & 4 & 2 & 1
\end{pmatrix}=(1\quad 3\quad 2\quad 4)$ is a cycle of length $5-1=4$.

But these two cases seem to be just exceptions.

For other (small) values of $p$ and every primitive root $g$ of $p$, there are no cycles of length $p-1$.

I can prove that $g=2$ and $g=\frac{p+1}{2}$ cannot produce such a big cycle, but I do not have any idea how to attack the problem for other values of $g$.

I conjecture that for every $g$ there is not such a big cycle (for sufficiently large $p$)

My question is:

Do only finitely many primes $p$ exist, such that for some primitive root $g$ the permutation $\sigma_g(p)=\begin{pmatrix} 1 & 2 & \ldots & {p-1} \\ g^1\pmod{p} & g^2\pmod{p} & \ldots & g^{p-1}\pmod{p} \end{pmatrix}$ is a cycle?