Are all totient values of Fibonacci Numbers distinct?

This question was inspired while I was seeing how certain recurrence relations would behave when I applied Multiplative Functions.

Let $F_{n}$ be a sequence for which $F_{1}=1,F_{2}=1$, and $F_{n}=F_{n-1}+F_{n-2}$. For any $n,m>8$, is it always true that $\phi(F_n)$ is not equal to $\phi(F_m)$, when $\phi$ is the Euler Totient Function?

A case when $n$ divides $m$ appeared be rather simple.

$n|m$ would imply $F_{n}|F_{m}$. Then let $F_{m}$ be $F_{n}$ multiplied by $k$.

$\phi(F_{m}) \ge \phi(F_{n})\phi(k)$.

$F_{n+1}=F_{n}+F_{n-1}<2F_{n}$, and $F_{n+2}=2F_{n}+F_{n-1}>2F_{n}$.

Thus $k$ may never be equal to 2. Therefore, $\phi(k)$ will always be greater than 1 implying $\phi(F_{m})=\phi(k)\phi(F_{n})>\phi(F_{n})$.

Therefore, for the case when $n$ divides $m$, $\phi(F_m)$ is always not equal to $\phi(F_n)$ . However, other cases proved to be rather difficult, such as proving $\phi(F_n) \neq \phi (F_{n+1})$.

Can anybody prove or disprove this question? Any Help would be greatly appreciated.

You have $\phi(5)=\phi(8)$, and $\phi(13)=\phi(21)$.

There is no really easy way, because the fibonacci numbers are the repunits of $j(-3),$ which means they are of the form $b^n+b^{-n}$, where $n=\frac 12(\sqrt{j+2}+\sqrt{j-2})$. This means that there is no particular rule that the period of a prime must divide the totient of the prime.