I'm not sure what to call this proposition in relation to Fermat's Little Theorem (it is not the converse, though it seems related), but I am interested to know if the following holds:
$$ a^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1} $$ for $p$ prime and $a \neq 1$.
If this does hold, what is the proof? If not, are there any conditions on $x$ required in order to have $a^x \equiv 1 \pmod{p}$?