$f(a)^{f(p)}\equiv f(a)\pmod p$ is satisfied if (but not only if) $f(p) \equiv 1 \pmod{p-1}$, i.e. $f(p) = c_p (p-1) + 1$. But there's no reason a priori why $c_p$ should be the same for all primes, or why $f(n)$ should be $c(n-1) + 1$ for non-prime $n$. I think you've only considered a negligible subset of the possible fractions $f$ satisfying the reduced property for $a=b$.

That's a good point. I'll look more into this. 🙂

Actually we can argue that for each prime $p$ there is a $c_p$ such that $f(p) = c_p(p-1) + 1$. If $f(n) \bmod p$ takes fewer than $p$ distinct values then by the pigeonhole principle there must be $a \not\equiv b$ for which $f(a) \equiv f(b)$ and we can derive a contradiction from $f(a)^{f(p)} \equiv f(a) \equiv f(b)$. Now, let $x$ be such that $f(x)$ is a generator of the multiplicative group modulo $p$; it's a cyclic group, so $f(x)^{f(p)} \equiv f(x)$ requires $f(p) \equiv 1 \pmod {p-1}$.

Then the original property is equivalent to the pair of properties: $\forall p \textrm{ prime}: f(a) \equiv f(b) \pmod p \iff a \equiv b \pmod p$ and $\forall p \textrm{ prime}: f(p) \equiv 1 \pmod {p-1}$. By Dirichlet's theorem on arithmetic progressions we have that for every prime $p$ there are an infinite number of primes $q = 1 + mp$. Then $f(q) \equiv 1 \pmod {mp}$, so $f(1) \equiv f(1 + mp) \equiv 1 \pmod {p}$. Thus $f(1) = 1$, since otherwise a prime factor of $f(1)$ would give a contradiction. $f(2) - 1$ is odd but cannot be divisible by an odd prime, so $f(2) = 2$. The only equivalence class for $f(3) \pmod 3$ is $0$ and both $f(3) - 1$ and $f(3) - 2$ must be $2$-smooth. Therefore $f(3) = 3$. Not sure how far this can be pushed.

It gets messy trying to push it. $f(5) \equiv 5 \pmod {12}$ follows from previous observations; $f(5) \in \{0, 4\} \pmod 5$ follows because the other equivalence classes are already accounted for. Then $f(5) - 1$ must be a power of two (it must be $5$-smooth and coprime to $3$ and $5$); $f(5) - 3$ must be a power of two (similarly); so $f(5) = 5$. But $f(7) \equiv 7 \pmod {30}$ and $f(7) \in \{0,4,6\} \pmod 7$ gives $f(7) = 5^u + 2 = 2^v + 5$, and while the technique used to prove Mihailescu's theorem might stretch to this, it's not looking promising as a proof strategy.

On the positive side, it's sufficient to show that $f(p) = p$ for all primes $p$. If so then for each $1 \le a < p$ there is a prime $q = a + mp$ by Dirichlet's theorem on arithmetic progressions: then $f(q) \equiv a \pmod p$ and since each equivalence class maps to a single equivalence class this means that in general $f(a) \equiv a \pmod p$. Thus in general $f(n) = n$, since otherwise we can find a prime $p > \max(n, f(n))$ and derive a contradiction.