For any suitable $n$ that has primitive roots (i.e. $n$ of the form $2, 4, p^j, 2p^j$, where $p$ is an odd prime), there exist primitive root(s). In the case that $n$ has more than one primitive root, how can I show that they don't generate $\mathbb{Z}^{\times}_n$ (the subset of $\mathbb{Z}_n$ whose elements are coprime to $n$) in the same order?
So, for $a$ and $b$ both distinct primitive roots, $a^k \neq b^k$, $k \in [1, \phi(n)-1]$.
The following is an argument that works, if I am interpreting "in the same order" correctly. Let $g$ and $h$ be primitive roots modulo $n$ and $i$ be a positive integer less than $\phi(n)$ such that $g^i \equiv h\pmod{n}.$ If the powers of $g$ and $h$ cycle through the coprime residues in the same order, then we have $$gh\equiv g^{i+1}\equiv h^2\pmod{n}.$$ Cancelling $h$ from both sides yields $g\equiv h\pmod{n},$ so $g$ and $h$ are not distinct modulo $n.$