Q1: Is it true that if $d$, if $d|p-1$, where $p$ a prime, then there for all $(x,p) = 1$, are exactly $d$ distinct remainders when $x^{\frac{p-1}{d}}$ is divided by $p$?
Q2: For what composite number $n$ does the above hold true?
Motivation: If this is true then $d = 1$ gives Fermat's Little Theorem.
Yes, this is true. By the existence of primitive roots modulo $p$, the integers modulo $p$ form a cyclic group of order $p-1$. (By Lagrange's theorem, knowing that it's a group already implies Little Fermat.)
We can now use the following general fact about a cyclic group $G$ of order $n$: for each divisor $d \mid n$, there is a unique subgroup of order $d$, and its elements are precisely the $n/d$th powers of elements of $G$.