This problem is beautiful, and is closely related to the Turkey'95'Q6. I will post my solution later, by the key ingredients are, the statement in the link I've provided, the primeness of $f(p)$, for any $p$, and, the relation between ${\rm gcd}(f(m),f(n))$ ,and, $f({\rm gcd}(m,n))$. EDIT : My proof is below: For the forward direction, suppose that, $f(mn)=f(m)f(n)$, and we want to prove that, ${\rm lcm}(f(m),f(n))=f({\rm lcm}(m,n))$. Noticing, ${\rm lcm}(m,n){\rm gcd}(m,n)=mn$, we have, $$ f(mn)=f(m)f(n)=f({\rm gcd}(m,n))f({\rm lcm}(m,n))={\rm gcd}(f(n),f(m)){\rm lcm}(f(n),f(m)) $$the statemet we want to prove, is equivalent to proving, ${\rm gcd}(f(n),f(m))=f({\rm gcd}(m,n))$. To prove this, we first make a claim: Claim: For any prime $p$, $f(p)$ is a prime. Proof: We argue as in the problem, posted in the link. We will first show that, $m\mid n\iff f(m)\mid f(n)$. First, suppose, $m\mid n$. Then, $n=mk$, therefore, $f(n)=f(mk)=f(m)f(k)\implies f(m)\mid f(n)$. Next, suppose, $f(m)\mid f(n)$, and that, $f(n)=kf(m)$, for some $k$. Since $f(\cdot)$ is surjective, there exists an $\ell$, such that, $f(\ell)=k$, and with this, we have, $f(n)=f(\ell)f(m)=f(m\ell) \implies n=m\ell$, since, $f(\cdot)$ is injective, Hence, $m \mid n \iff f(m)\mid f(n)$. With this in place, let, $n=p$ for a prime $p$, and let $q$ be any divisor of $f(p)$. Clearly, $q\mid f(p) \implies q\mid p$, hence, $q=1$, or, $q=p$. $q=1$ corresponds to $1$ -since, $f(1)=f(1)^2 \implies f(1)=1$, hence, $f(p)$ must be a prime. Having proven the lemma, we are now almost done with the first part. Let, $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$, and, $m=p_1^{\beta_1}\cdots p_k^{\beta_k}$ (here, consider, $\{p_1,\dots,p_k\}$ as the union of all primes, dividing at least one of $n,m$, thus, exponents can very well be 0, but not simultaneously). We are to prove, $$ f({\rm gcd}(m,n))=f\left(p_1^{\theta_1}\cdots p_k^{\theta_k}\right) \overbrace{=}^{?} {\rm gcd}\left(f(p_1)^{\alpha_1}\cdots f(p_k)^{\alpha_k},f(p_1)^{\beta_1}\cdots f(p_k)^{\beta_k}\right), $$where, $\theta_i = \min \{\alpha_i,\beta_i\}$. Using now, the fact that, $f(p_i)$'s are all primes, the right hand side above, is, $$ f(p_1)^{\theta_1}\cdots f(p_k)^{\theta_k}, $$which, by assumption, is equal to, $f(p_1^{\theta_1}\cdots p_k^{\theta_k})$. Hence, we are done with $\implies$ part. For the other part, we will again, begin by proving, $$ {\rm lcm}(f(m),f(n))=f({\rm lcm}(m,n))\implies \left(m \mid n \iff f(m)\mid f(n)\right). $$To see this, suppose, $m\mid n$, and that, $n=mk$. We have, $$ f({\rm lcm}(m,n))=f(n)={\rm lcm}(f(n),f(m))\implies f(m)\mid f(n). $$For the other part, notice, if, $f(m)\mid f(n)$, then, $f(n)=kf(m)=f(\ell)f(m)$, for some $k,\ell$, with, $f(\ell)=k$ (as above), hence, $$ {\rm lcm}(f(m),f(n))=f(n)=f({\rm lcm}(m,n))\implies n = {\rm lcm}(m,n) \implies m\mid n, $$where, the last equality above follows again from the fact that, $f(\cdot)$ is injective. Notice that, as a consequence, $f(1)=1$ (we can't now, rely on $f(mn)=f(m)f(n)$ to deduce $f(1)=1$, as that is what we are trying to prove). Now, to conclude, recall that, under what has just been proven, $f(p)$ is a prime, and suppose, $m=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$, and, $n=p_1^{\beta_1}\cdots p_k^{\beta_k}$. We will now, prove, $f(mn)=f(m)f(n)$, for $(m,n)=1$. To see this, note that, if $q\mid f(m),f(n)$, then, letting, $q=f(i)$ (using surjectivity), we have, $f(i)\mid f(m) \implies i\mid m$, and similarly, $f(i)\mid f(n)\implies i \mid n$. Hence, $i\mid (m,n)$, and thus, $i=1$, namely, $f(m)$ and $f(n)$ are coprime. Therefore, ${\rm lcm}(f(m),f(n))=f(m)f(n)=f({\rm lcm}(m,n))=f(mn)$. Using this, we get, $$ f\left(p_1^{\alpha_1+\beta_1}\cdots p_k^{\alpha_k+\beta_k}\right)=f\left(p_1^{\alpha_1+\beta_1}\right)\cdots f\left(p_k^{\alpha_k+\beta_k}\right). $$In order to finish the proof, it suffices to show, $f(p^t)=f(p)^t$, for every prime $p$, and positive integer $t$. For this, you can either refer to the argument in PEN, or follow mine, as below. We will argue, as follows. First, $p^i \mid p^j$, for, $i<j$, implies, $f(p^i)\mid f(p^j)$. Hence, $f(p)<f(p^2)<f(p^3) <\cdots$. Now, consider, $f(p^2)$. If, $f(i)\mid f(p^2)$, then, $i\mid p^2$, hence, $i\in \{1,p,p^2\}$. We now prove that, $f(p^2)=f(p)^t$ for some $t$. To see this, suppose that $f(p^2)=f(p)f(q)$, for some $q$ (well, $f(p)<f(p^2)$, and, since $f(p)\mid f(p^2)$, such $q$ must exist). Then, $f(q)\mid f(p^2)\implies q\mid p^2 \implies q\in\{1,p,p^2\}$. Clearly, $q\neq p^2$, and, $q\neq 1$. Hence, $q=p$. Going like this, we must have that, $f(p^i)=f(p)^k$. The final thing, is to clinch the loose end, by showing that, $i=k$. Suppose that, $f(p^i)=f(p)^i$, for $i=1,2,\dots,k-1$, but, jumps for $f(p^k)$, namely, $f(p^k)=f(p)^i$, for $i>k$. Now, consider, $N=f(p)^k$. Writing, $N=f(a)$ for some $a$, implies that,$f(a)=f(p)^k$. In particular, $p\mid a$. Moreover, if $a$ had a prime divisor $r$, other than $p$, we would have, $f(r)\mid f(a)=f(p)^k$. However, $f(p)^k$ is a prime power, and since, $f(r) \neq f(p)$, and, $f(r)$ is also a prime, this is impossible. Namely, $a=p^{\ell}$. However, it is now very clear that, such an $\ell$ cannot exist (due to, $\ell \geq k$, but, since $f(p^{\ell})<f(p^k)$, we have, $\ell <k$, a contradiction). We are done.