Primitive roots as roots of equations. 
Take $g$ to be a primitive root $\pmod p$, and $n \in \{0, 1,\ldots,p-2\}$ write down a necessary sufficient condition for $x=g^n$ to be a root of $x^5\equiv 1\pmod p$ . This should depend on $n$ and $p$ only, not $g$.
How many such roots $x$ of this equation are there?  This answer may only depend on $p$.

At a guess for the first part I'd say as $g^{5n} \equiv g^{p-1}$ it implies for $x$ to be a root $5n \equiv p-1 \pmod p$.  No idea if this is right and not sure what to do for second part.  Thanks for any help.

Hint. In any abelian group, if $a$ has order $n$, then $a^r$ has order $n/\gcd(n,r)$. 
(Your idea is fine, except that you got the wrong congruence: it should be $5n\equiv p-1\pmod{p-1}$, not modulo $p$; do you see why?)
For the second part, you'll need to see what you get from the first part. That will help you figure it out.
