Concerning Primitive roots and exponents.

Question

If for any $u$,such that 1 < u < p; given $\bmod p$; if $m^u \equiv u V \pmod{p}$ where $p$ does not divide $(V - 1)$ for any $V$ then $m$ is not a primitive root. Is this true?

Answer 1

Based on the comments, the question appears to be this: "If $m$ is a primitive root modulo a prime $p$ then for any integer $1 \le u < p$ we have $m^u \equiv s \pmod{p}$ where $1 \le s < p$."

Since $s$ can be represented by a residue system between $1 \le s < p$, this is true for any $m$ not congruent to $0$ modulo $p$ and so it is also true for primitive roots.