Define g(m) = \frac{f(m)}{m}
g(m) = 1 iff m = 1, (using iii)
p\mid f(p) for any prime p, (using i), and then g(p) \in \mathbb{N} for prime p
From (ii) we have that g(mn)g(d) = g(m)g(n) \text{ if } \gcd(m,n) = d
Using this, we get that:
(1) g(mn) = g(m)g(n) \text{ if } \gcd(m,n) = 1
(2) g(md) = g(m) \text{ if } d\mid m
g(p^s) = g(p^{s-1}\cdot p) = g(p^{s-1}) for any s\ge 2 \text{ (using (2))}, and from this follows that g(p^s) = g(p)
Then g(\prod p_i^{a_i}) = \prod g(p_i) \text{ (using (1))}
Now we need to find the value of g(p) for p prime. We will prove that it must be a power of p
pf(f(p)) = f(p)^2 \Rightarrow g(p\cdot g(p)) = g(f(p)) = \frac{f(f(p))}{f(p)} = \frac{f(p)}{p} = g(p)
If q\mid g(p) for some prime q\ne p then g(p)g(q) \mid g(p\cdot g(p)) = g(p) since p and q are prime divisors of p\cdot g(p). This implies g(q) = 1 which is a contradiction. Then g(p) = p^{v_p} for some v_p, and since g(p) \ne 1 we have that v_p\ge 1.
Write m = \prod p_i^{a_i}. Then f(m) = m\cdot g(m) = m\prod g(p_i) = m\prod p_i^{v_{p_i}}
And we are done, because this satisfies the conditions of the problem