For $n$ coprime with $3$ define $v(n)$ as the number such that $n\equiv 2^{v(n)}\mod{27}$, and if $(n,3)>1$ define $v(n)=0$. With this we have $v(n)$ is periodic with least period $27$. \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 & \cdots \\ \hline v(n) & 18 & 1 & 0 & 2 & 5 & 0 & 16 & 3 & 0 & 6 & 13 & 0 & 8 & 17 & 0 & 4 & 15 & 0 & 12 & 7 & 0 & 14 & 11 & 0 & 10 & 9 & 0 &\cdots\\ \hline \end{array}
Now for each $m\geq 1$ define $f_{m}(n):=m\cdot v(n)\mod{18}$. I've made some calculations and I found that $f_{m}$ has least period $27$ if $(m,3)=1$, also $f_{3}$, $f_{6}$, $f_{12}$ and $f_{15}$ have least period $9$ and finally $f_{9}$ has least period $3$.
My question is: How can you find the least period of $f_{m}$ as a formula in terms of $m$? (without explicitly calculating all the values). And in general, if $g$ is a primitive root mod $p^{k}$, $v(n)$ is such that $n\equiv g^{v(n)}\mod{p^{k}}$, and define $f_{m}(n):=m\cdot v(n)\mod{\varphi(p^{k})}$. How can we find the least period of each $f_{m}$?
Edit: Additional info:
\begin{array}{|c|c|c|c|} \hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 & 27 & \cdots \\ \hline v(n) & 18 & 1 & 0 & 2 & 5 & 0 & 16 & 3 & 0 & 6 & 13 & 0 & 8 & 17 & 0 & 4 & 15 & 0 & 12 & 7 & 0 & 14 & 11 & 0 & 10 & 9 & 0 &\cdots\\ \hline f_{1}(n) & 0 & 1 & 0 & 2 & 5 & 0 & 16 & 3 & 0 & 6 & 13 & 0 & 8 & 17 & 0 & 4 & 15 & 0 & 12 & 7 & 0 & 14 & 11 & 0 & 10 & 9 & 0 & \cdots \\ \hline f_{2}(n) & 0 & 2 & 0 & 4 & 10 & 0 & 14 & 6 & 0 & 12 & 8 & 0 & 16 & 16 & 0 & 8 & 12 & 0 & 6 & 14 & 0 & 10 & 4 & 0 & 2 & 0 & 0 & \cdots \\ \hline f_{3}(n) & \color{red}{0} & \color{red}{3} & \color{red}{0} & \color{red}{6} & \color{red}{15} & \color{red}{0} & \color{red}{12} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{3} & \color{cyan}{0} & \color{cyan}{6} & \color{cyan}{15} & \color{cyan}{0} & \color{cyan}{12} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{3} & \color{red}{0} & \color{red}{6} & \color{red}{15} & \color{red}{0} & \color{red}{12} & \color{red}{9} & \color{red}{0} & \cdots \\ \hline f_{4}(n) & 0 & 4 & 0 & 8 & 2 & 0 & 10 & 12 & 0 & 6 & 16 & 0 & 14 & 14 & 0 & 16 & 6 & 0 & 12 & 10 & 0 & 2 & 8 & 0 & 4 & 0 & 0 & \cdots \\ \hline f_{5}(n) & 0 & 5 & 0 & 10 & 7 & 0 & 8 & 15 & 0 & 12 & 11 & 0 & 4 & 13 & 0 & 2 & 3 & 0 & 6 & 17 & 0 & 16 & 1 & 0 & 14 & 9 & 0 & \cdots \\ \hline f_{6}(n) & \color{red}{0} & \color{red}{6} & \color{red}{0} & \color{red}{12} & \color{red}{12} & \color{red}{0} & \color{red}{6} & \color{red}{0} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{6} & \color{cyan}{0} & \color{cyan}{12} & \color{cyan}{12} & \color{cyan}{0} & \color{cyan}{6} & \color{cyan}{0} & \color{cyan}{0} & \color{red}{0} & \color{red}{6} & \color{red}{0} & \color{red}{12} & \color{red}{12} & \color{red}{0} & \color{red}{6} & \color{red}{0} & \color{red}{0} & \cdots \\ \hline f_{7}(n) & 0 & 7 & 0 & 14 & 17 & 0 & 4 & 3 & 0 & 6 & 1 & 0 & 2 & 11 & 0 & 10 & 15 & 0 & 12 & 13 & 0 & 8 & 5 & 0 & 16 & 9 & 0 & \cdots \\ \hline f_{8}(n) & 0 & 8 & 0 & 16 & 4 & 0 & 2 & 6 & 0 & 12 & 14 & 0 & 10 & 10 & 0 & 14 & 12 & 0 & 6 & 2 & 0 & 4 & 16 & 0 & 8 & 0 & 0 & \cdots \\ \hline f_{9}(n) & \color{red}{0} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{9} & \color{red}{0} & \cdots \\ \hline f_{10}(n) & 0 & 10 & 0 & 2 & 14 & 0 & 16 & 12 & 0 & 6 & 4 & 0 & 8 & 8 & 0 & 4 & 6 & 0 & 12 & 16 & 0 & 14 & 2 & 0 & 10 & 0 & 0 & \cdots \\ \hline f_{11}(n) & 0 & 11 & 0 & 4 & 1 & 0 & 14 & 15 & 0 & 12 & 17 & 0 & 16 & 7 & 0 & 8 & 3 & 0 & 6 & 5 & 0 & 10 & 13 & 0 & 2 & 9 & 0 & \cdots \\ \hline f_{12}(n) & \color{red}{0} & \color{red}{12} & \color{red}{0} & \color{red}{6} & \color{red}{6} & \color{red}{0} & \color{red}{12} & \color{red}{0} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{12} & \color{cyan}{0} & \color{cyan}{6} & \color{cyan}{6} & \color{cyan}{0} & \color{cyan}{12} & \color{cyan}{0} & \color{cyan}{0} & \color{red}{0} & \color{red}{12} & \color{red}{0} & \color{red}{6} & \color{red}{6} & \color{red}{0} & \color{red}{12} & \color{red}{0} & \color{red}{0} & \cdots \\ \hline f_{13}(n) & 0 & 13 & 0 & 8 & 11 & 0 & 10 & 3 & 0 & 6 & 7 & 0 & 14 & 5 & 0 & 16 & 15 & 0 & 12 & 1 & 0 & 2 & 17 & 0 & 4 & 9 & 0 & \cdots \\ \hline f_{14}(n) & 0 & 14 & 0 & 10 & 16 & 0 & 8 & 6 & 0 & 12 & 2 & 0 & 4 & 4 & 0 & 2 & 12 & 0 & 6 & 8 & 0 & 16 & 10 & 0 & 14 & 0 & 0 & \cdots \\ \hline f_{15}(n) & \color{red}{0} & \color{red}{15} & \color{red}{0} & \color{red}{12} & \color{red}{3} & \color{red}{0} & \color{red}{6} & \color{red}{9} & \color{red}{0} & \color{cyan}{0} & \color{cyan}{15} & \color{cyan}{0} & \color{cyan}{12} & \color{cyan}{3} & \color{cyan}{0} & \color{cyan}{6} & \color{cyan}{9} & \color{cyan}{0} & \color{red}{0} & \color{red}{15} & \color{red}{0} & \color{red}{12} & \color{red}{3} & \color{red}{0} & \color{red}{6} & \color{red}{9} & \color{red}{0} & \cdots \\ \hline f_{16}(n) & 0 & 16 & 0 & 14 & 8 & 0 & 4 & 12 & 0 & 6 & 10 & 0 & 2 & 2 & 0 & 10 & 6 & 0 & 12 & 4 & 0 & 8 & 14 & 0 & 16 & 0 & 0 & \cdots \\ \hline f_{17}(n) & 0 & 17 & 0 & 16 & 13 & 0 & 2 & 15 & 0 & 12 & 5 & 0 & 10 & 1 & 0 & 14 & 3 & 0 & 6 & 11 & 0 & 4 & 7 & 0 & 8 & 9 & 0 & \cdots \\ \hline \end{array}
You are interested in finding the least period $q$ such that, for all $n$,
$mv(n)\equiv mv(n+q)\pmod {18}$
$\Leftrightarrow 2^{mv(n)}\equiv 2^{mv(n+q)}\pmod {27}$
$\Leftrightarrow n^m\equiv (n+q)^m\pmod {27},$ providing $n$ and $n+q$ are coprime to $3$.
The period has to be a factor of $27$ and so $q\in \{ 1,3,9,27 \}$.
Using the Lemma given below, the period of $f_{m}(n)$ is $\frac {27}{(m,27)}$ for $1\le m\le 17$.
The general case for $p$ odd
The crucial equation is $ n^m\equiv (n+q)^m\pmod {p^k}.$
For $1\le m<\varphi(p^{k})$ the period is $\frac {p^k}{(m,p^k)}$.
Lemma (Hensel lifting)
For an odd prime $p$, if $a$ and $b$ are coprime to $p$ and $a\equiv b\pmod p$, then the power of $p$ dividing $a^m-b^m$ is the sum of the powers of $p$ dividing $a-b$ and $m$.