Corollary of Fermat's Little Theorem to $a^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1}$ 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 $a^x \equiv 1 \pmod{p}$?

For each prime $p$, there exists a primitive root $r$ which has the property you want: $$r^x \equiv 1 \pmod{p} \implies x \equiv 0 \pmod{p-1}.$$
For example, if we work modulo $7$, $3$ is a primitive root, because
\begin{align}
   3^0 &\equiv 1 \pmod 7 \\
   3^1 &\equiv 3 \pmod 7 \\
   3^2 &\equiv 2 \pmod 7 \\
   3^3 &\equiv 6 \pmod 7 \\
   3^4 &\equiv 4 \pmod 7 \\
   3^5 &\equiv 5 \pmod 7 \\
   3^6 &\equiv 1 \pmod 7 \\
\end{align}
so the powers of $3$ have a period of exactly $6$. It follows that $2$ will not work mod $7$, because $$2^3 \equiv (3^2)^3 \equiv 3^6 \equiv 1 \pmod 7.$$
This MSE post gives a short proof of the existence of primitive roots modulo any prime $p$.
