If $a^x \equiv a^y \pmod{m}$, does this imply $x \equiv y \pmod{\varphi(m)}$? I know that $x \equiv y \pmod{\varphi(m)} \implies a^x \equiv a^y \pmod{m}$, so I was wondering if the converse also holds. If it does not hold in general, then are there any conditions where it does (such as assuming $(a, m)=1$)?
*Here $\varphi$ denotes the Euler totient function.
Edit: I see that the answer is no. However, the reason I was wondering about this is that I was watching this (https://youtu.be/f1oO9dEkqso?t=1386) lecture and don't know why else would Prof. Borcherds claim that $a_i \equiv 1 \pmod{p_{i}^{k_i}}$ is solvable if $\varphi(p_{i}^{k_i}) \mid n$ if the property in question is not true. In other words, how else is what's written justified?

take $m=8$ and $a=2$, so $\phi(m)=4$ and $2^{3}\equiv2^{4}\equiv0(mod8)$ yet $3\not \equiv4(mod4)$
