If $x^d \equiv 1 \pmod{p}$ has full number of roots, $d\mid p-1$? Let $p$ be a prime.
What I know is that $d\mid p-1$ then $x^d \equiv 1 \pmod{p}$ has the maximum possible number of roots, that is, $d$.
I am wondering if the converse holds. 
I have searched for $d\not\mid p-1$ that still has $d$ many roots in the equation, but I could not find one up to $p=23$ (unless my calculation was not complete). Is there a counter-example to the converse?

Use Discrete Logarithm
$$x^d\equiv1\pmod p\implies$$ 
$d\cdot $ind$_gx\equiv0\pmod{p-1}$ for an primitive root $g\pmod p$
What if ind$_gx\equiv1\pmod{p-1}$
Alternatively,  let $G=(d,p-1)$  and $\dfrac dD=\dfrac{p-1}Q=G\implies(D,Q)=1$
$D\cdot $ind$_gx\equiv0\pmod Q\implies Q$ must divide ind$_gx$
For maximum root $Q$ must be $=1$
