TITLE: Is the identity map the only map from the positive integers to itself that simultaneously preserves multiplication, order, and primes?
QUESTION [9 upvotes]: Is the identity map the only map $f$ from the positive integers $\mathbb{Z}_{+}$ to itself that satisfies all three of the following properties?

$f(mn)=f(m)f(n)$ for all positive integers $m$ and $n$ (which implies that $f(1)=1$)
$f(m)<f(n)$ whenever $m<n$ (which implies that $f$ is injective)
$f(p)$ is prime for any prime number $p$

The above three properties (actually, for property 2, the weaker assumption that $f$ is injective suffices) imply that applying $f$ to any positive integer will not change the prime signature, because all that happens with $f(n)$ is that each prime $p$ dividing $n$ is replaced with $f(p)$ while maintaining the same exponent.
Here are some examples that suggest that the answer might be "Yes" (because they only satisfy properties 1 and 3 and injectivity, but are not order-preserving on all positive integers, although they are order-preserving on prime numbers):

Let $f(p)$ be the next prime following $p$, with $f$ extended to composite numbers using (complete) multiplicativity. Then, $8<9$, while $f(8)=f(2^3)=f(2)^3=3^3=27>25=5^2=f(3)^2=f(3^2)=f(9)$.
Let $f$(the $n$th prime) be the $(2n)$th prime. Then, $27<32$, while $f(27)=f(3^3)=f(3)^3=7^3=343>243=3^5=f(2)^5=f(2^5)=f(32)$.
Let $f(p)$ be the $p$th prime. Then, $f(8)=27>25=f(9)$, as in the first example.
Let $f$(the $n$th prime) be the $(n^2)$th prime. Then, $3<4$, while $f(3)=7>4=2^2=f(2)^2=f(2^2)=f(4)$.

But of course, specific examples (using inductive reasoning) are not enough to conclude that the answer is in fact "Yes".

REPLY [7 votes]: Building on the discussion in the comments of Teresa's answer, this follows from the six exponentials theorem. I will try to write a self-contained answer here.
First, as you say the given conditions imply that $f$ restricts to some strictly monotonic function on the primes, and is determined by its values on the primes. If $p_1, \dots p_k$ is any finite set of distinct primes, then strict monotonicity of $f$ means that if $\prod p_i^{e_i} < \prod p_i^{f_i}$ then $\prod f(p_i)^{e_i} < \prod f(p_i)^{f_i}$. Taking logarithms and subtracting, this gives
$$\sum q_i \log p_i < 0 \Rightarrow \sum q_i \log f(p_i) < 0, q_i \in \mathbb{Q}.$$
In other words, $f$ induces a strictly order-preserving linear map $F$ from the $\mathbb{Q}$-subspace of $\mathbb{R}$ spanned by $\{ \log p_i \}$ to the $\mathbb{Q}$-subspace of $\mathbb{R}$ spanned by $\{ \log f(p_i) \}$. This implies that $F$ is continuous (with respect to the subspace topology coming from $\mathbb{R}$), which gives that it must in fact be given by multiplication by some fixed real number $r$. Said another way:

Corollary: For real coefficients $r_i$, we have $\sum r_i \log p_i = 0 \Leftrightarrow \sum r_i \log f(p_i) = 0$. Equivalently, there is some $r$ such that $\log f(p_i) = r \log p_i$ for all primes $p_i$.

This is quite a strong condition. It's equivalent to the condition that $\frac{\log f(p)}{\log p} = \frac{\log f(q)}{\log q}$ for distinct primes $p, q$, which assuming Schanuel's conjecture (which implies that the logarithms of the primes are algebraically independent) gives $f(p) = p, f(q) = q$. (We can also assume the weaker four exponentials conjecture instead.)
But actually we can avoid Schanuel's conjecture by considering three distinct primes $p, q, r$ instead. The six exponentials theorem implies that if $f(p) \neq p$ then (using that unique prime factorization implies that the logarithms of the primes are linearly independent over $\mathbb{Q}$) the matrix
$$\left[ \begin{array}{cc} \log p & \log q & \log r \\ \log f(p) & \log f(r) & \log f(r) \end{array} \right]$$
has rank $2$. But we showed above that the second row must be a multiple of the first, so this matrix has rank $1$; taking contrapositives, it follows that $f(p) = p$, and since the argument works for any three distinct primes we get $f(p) = p$ for all primes $p$.