This question has bugged me for a long time. I've asked some of my professors and they seem to believe that these objects haven't been studied. I'm prone to believe that the construction is too simple to not have been looked at in the cosmos of mathematics publications.
Now the construction is simple. Let us consider completely multiplicative bijective functions on $\mathbb{N}$. That is if $q$ is such a function, then $q(mn) = q(m)q(n)$ and for all $m$ there exists an $n$ such that $q(n) = m$. Now lets extend this function to $\mathbb{Z}$ by letting $q(-m) = - q(m)$ and $q(0) = 0$. From this there is an obvious extension to $\mathbb{Q}$ by letting $q(m/n) = q(m)/q(n)$.
Here is where we start fiddling with things. Let us define
$$a \oplus_{q} b = q(q^{-1}(a) + q^{-1}(b))$$ which paired with multiplication turns $\{\mathbb{Q},\oplus_q,\cdot\}$ into a field. Call this field/space $\mathbb{Q}_q$. Now in some instances extending $q$ from $\mathbb{Q}$ to a larger domain is perfectly possible. Remembering these functions are simply permutations of prime numbers we can consider the following.
Let $q$ be such that $q(2) = 3$ and $q(3) = 2$ and $q(p) = p$ for all other $p$ prime. Instantly we arise at the following little conclusion
$$q(\sqrt{3}) = \pm \sqrt{2}$$
which follows because
$$2 = q(3) = q(\sqrt{3}\sqrt{3}) = q(\sqrt{3})q(\sqrt{3}) = (\pm\sqrt{2})(\pm\sqrt{2})$$
To be simple we will restrict $q$ to take positives to positives so that $q(\sqrt{3}) = \sqrt{2}$. The more general result is simpler to write out, if $\{p_i\}_{i=1}^{\infty}$ is the sequence of primes and $\{b_i\}_{i=1}^\infty$ is a sequence of rationals finitely many nonzero.
$$q(\prod_{i=1}^\infty p_i^{b_i}) = \prod_{i=1}^\infty q(p_i)^{b_i}$$
Which extends our definition of $q$ to a much larger domain. But sadly, not all of $\bar{\mathbb{Q}}$. However, there is a manner of tricking the definition to work.
Taking polynomials $\mathbb{Q}_q[x]$ and modding out by some polynomial $f$ in $\mathbb{Q}_q[x]$ produces a field isomormphic to some field extension of $\mathbb{Q}$--the same as how $\mathbb{Q}_q(\sqrt{3}) \simeq \mathbb{Q}(\sqrt{2})$. Which sends a set of algebraic conjugates to algebraic conjugates arbitrarily ($\{\sqrt{3},-\sqrt{3}\} \to \{\sqrt{2},-\sqrt{2}\}$). Which seems to imply that there is an extension of $q$ to $\bar{\mathbb{Q}}$.
What I thought of adding from here is rather intuitive, let us define a metric on this field. The metric will be simple enough:
$$|a|_q = q^{-1}(|a|)$$
which satisfies the definition of a metric between the spaces $\mathbb{Q}_q \to \mathbb{Q}^+$.
$$|ab|_q = |a|_q|b|_q$$ $$|a \oplus_q b|_q = |a|_q + |b|_q$$ $$|a|_q = 0 \Rightarrow a = 0$$
Now what is interesting, and the crux of what my question relies on, is taking Cauchy sequences under this metric. It follows cauchy sequences that converge under the usual metric (say $a_n \to e$) converge under this metric ($q(a_n) \to q(e))$ or they converge to something in $\hat{\mathbb{C}}$ for example. Which finally gives the idea that $q: \mathbb{C} \to \mathbb{C}$.
My question is rather simple with all of this knowledge. Is there any manner of explicitly solving for $q(x)$ for any $x$. I imagine with algebraic numbers it isn't impossible as
$$f(x) = a_0 \oplus a_1x \oplus a_2x^2 \oplus... \oplus a_nx^n = q(q^{-1}(a_1) + q^{-1}(a_2)y + ... + q^{-1}(a_n)y^n) = q(g(y))$$ so that $$\mathbb{Q}_q[x]/f(x) \simeq\mathbb{Q}[y]/g(y)$$ Implying that every root of $f$ is sent to a root of $g$. However it bugs me to wonder about something like $q(e)$, where the only expression I can think of for it is
$$1 \oplus 1 \oplus \frac{1}{q(1)q(2)} \oplus \frac{1}{q(1)q(2)q(3)} \oplus \frac{1}{q(1)q(2)q(3)q(4)} \oplus...$$
I'm asking for references to similar ideas, perhaps nuanced corrections to what I've written (I was a little hand waive-y writing this). And most especially if we can even call $q(\mathbb{R})$ the real line. Or if it is just some isomorphic object, wherein the idea of $q(e)$ equaling a number is gibberish, if instead it is just some abstract entity which when coupled with $\mathbb{Q}$ produces $\mathbb{R}$.
So to be forward, does this idea already exist, and if so what is the subject called? Is $q(e)$ a number in $\mathbb{R}$? Is $q(\mathbb{R})$ just some object topologically and algebraically equivalent to $\mathbb{R}$ or is it in fact $\mathbb{R}$?