How come that the number of distinct remainders $a_{k}$ for $g^{k}\equiv a_{k} \mod (n)$ for specific positive $n$ and any positive $g$ and $k=1,2,3...$ is never greater than $\varphi (n)$ (Euler's totient function)?
Example $n=6$:
For $g=2$ $$2^{1} \equiv 2 \mod (6)$$ $$2^{2} \equiv 4 \mod (6)$$ $$2^{3} \equiv 2 \mod (6)$$ For $g=3$ $$3^{1} \equiv 3 \mod (6)$$ $$3^{2} \equiv 3 \mod (6)$$ For $g=4$ $$4^{1} \equiv 4 \mod (6)$$ $$4^{2} \equiv 4 \mod (6)$$ For $g=5$ $$5^{1} \equiv 5 \mod (6)$$ $$5^{2} \equiv 4 \mod (6)$$ $$5^{3} \equiv 5 \mod (6)$$ For $g=11$ $$11^{1} \equiv 5 \mod (6)$$ $$11^{2} \equiv 1 \mod (6)$$ $$11^{3} \equiv 5 \mod (6)$$ $$...$$
The number of distinct remainders is never greater than $\varphi(6) = 2$.
Using the Chinese remainder theorem and the fact that $\varphi$ is multiplicative, you need only consider the case when $n = p^{t}$ is the power of a prime $p$.
In this case, if $p \nmid g$, the result is clear, as $g$ lies in the group of invertible elements of $\mathbb{Z} / p^{t} \mathbb{Z}$, of size $\varphi(p^{t})$.
If $p \mid g$, then $g^{t} = 0$ in $\mathbb{Z} / p^{t} \mathbb{Z}$, so the number of distinct remainders $g^{k}$ is at most $$t \le p^{t-1} \le (p -1) p^{t-1} = \varphi(p^{t}),$$ as $$p^{t-1} = (1 + (p-1))^{t-1} \ge 1 + (t-1)(p-1) \ge 1 + t - 1 = t.$$