$f(1)=1$. It's easy to conclude by induction, that $f(p^n)=p$, where $p$-prime. Let's consider prime factorization of $n$ and let $p_1,p_2,\ldots,p_s$ be set of its prime divisors. Then, $\prod_{i=1}^{a_1}f(p_1)^{i}\prod_{i=1}^{a_2}f(p_2)^{i}\ldots \prod_{i=1}^{a_s}f(p_s)^{i}\cdot A =n$, where $A$ is the multiple of rest divisors. So, $A = 1$, which means, each member of $A$ is equal to $1$. Answer is $f(p^a)=p$ for all primes $p$, and $f(x)=1$ for the rest.

It's obvious that at most one $f$ works. Now observe that $f(n) = e^{\Lambda(n)}$ works due to the identity $$\sum_{d\mid n} \Lambda(n) = \log n.$$($\Lambda$ is von Mangoldt function)

$P(1): f(1)=1$ $P(p): f(p)f(1)=p =>f(p)=p$, for primes $p$. Say that $f(p)=f(p^2)=...=f(p^{\alpha})=p$. $P(p^{\alpha+1}): f(1)f(p)f(p^2)...f(p^{\alpha})f(p^{\alpha+1})=p^{\alpha+1} =>f(p^{\alpha+1})=p$. Let $n = p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k}$. Notice that $p_i,p_i^2,...,p_i^{\alpha_i}$ are all divisors of $n$, thus: $P(n): p_1^{\alpha_1}p_2^{\alpha_2}...p_k^{\alpha_k} \cdot$ ( product of $f(m_i)$ where $m_i$ are all divisors of $n$ not of the form $p_i^t$ )$=p_1^{\alpha_1}...p_k^{\alpha_k}$. Hence $f(p^{\alpha})=p$, where $p$ is prime, and $f(m)=1$, where $m \neq p^{\alpha}$. Cute concept of a problem, although playing with divisors in FEs is certainly not new, it's implementation here seems quite cool and original...

Möbius inversion formula gives $$f(n)=\prod \limits _{d \mid n}d^{\mu(\frac{n}{d})}$$

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It's not uncommon for olympiad problems to be reused. There was this one problem that was used by the Soviets, South Korea, the Morocco TST, the Putnam, etc.

By condition we get that $f(1)=1$ since $1$ is the only divisor of $1$. Next, we get that $f(p)f(1)=p$ so $f(p)=p$. Also $f(p^2)f(p)f(1)=p^2$ so $pf(p^2)=p^2$ so $f(p^2)=p$. By induction for every integer $k$, we have that $f(p^k)=p$. By prime factorization we have that $n$ $=$ $\prod_{i=1}^{s}{p_{i}^{a_i}}$, and $P$ $=$ $\prod_{k=1}^{a_1}{f(p_{1}^{k})}$ $\cdots$ $\prod_{k=1}^{a_s}{f(p_{s}^{k})}$ times all $f(x)$ such that $x$ has at least two prime factors is equal to $n$ .But $P=\prod_{i=1}^{s}{p_{i}^{a_i}}=n$ since $f(p^k)=p$ , so for every $x$ such that it has at least two prime factors $f(x)=1$. So our solution is: $ \begin{cases} f(p^k)=p & \forall k \in \mathbb{Z}\ \text{and}\ \forall p \in \mathbb{P}\ \\ f(x)=1\ & \text{x has at least two prime factors} \\ \end{cases} $

We claim that $f(1)=1, f(p^k)=1$, and if $m$ isn't a prime power then $f(m)=1$. This clearly satisfies the relation for all $n$. Let $P(n)$ be the condition. Note that $P(1)$ gives $f(1)=1$. Then, we claim by induction that $f(p^k)= p$ for all $k\geq 1$. The base case is clear, $P(p)$ gives $f(p) = f(p)\cdot f(1) = p$. For the inductive step, considering $P(p^{k+1})$ gives $f(p^{k+1}) = \frac{p^{k+1}}{\prod_{i=0}^k f(p^k)} = p$. Now, we claim that all other values are 1. Consider some integer $m= \prod_{i=1}^t p_i^{e_i}$ which is not a prime power. Then, since $f\colon \mathbb{N}\to \mathbb{N}$, we have that based on $P(m)$: \[f(m) \mid \frac{m}{\prod_{i=1}^t f(p_i^{e_i})} = \frac{m}{m} = 1\]Thus, $f(m)=1$ if $m$ isn't a prime power and we're done.

With $n=1$ we find $f(1)=1$. Setting $n=p$ prime, we have $f(p)=p$. Now we prove by strong induction that $f(p^k)=p$ for $k\in\mathbb N$, with the base case already shown. If $f(p^i)=p$ for all $i<k$, then: $$p^k=\prod_{i=1}^kf(p^i)=p^{k-1}\cdot f(p^k)$$so $f(p^k)=p$. Now let $n$ not be a prime power. Writing $n$ as $\prod_{i=1}^kp_i^{a_i}$ (the primes $\{p_\ell\mid 1\le\ell\le k\}$ are not necessarily distinct and $k>1$), we have: $$\prod_{i=1}^kp_i=\prod_{i=1}^k\prod_{j=1}^{a_i}f(p_i^j)\cdot\prod_{S\subseteq\{p_\ell\mid 1\le\ell\le k\},1<|S|<n}f\left(\prod_{d\in S}d\right)\cdot f(n).$$Since: $$\prod_{i=1}^k\prod_{j=1}^{a_i}f(p_i^j)=\prod_{i=1}^k\prod_{j=1}^{a_i}p_i=\prod_{i=1}^kp_i^{a_i}$$the rest of the factors, including $f(n)$, must be $1$. So $f(p^k)=p$ for prime $p$ and $k\in\mathbb N$, and otherwise $f(n)=1$.

How to prove uniqueness？