What kinds of theories of real multiplication with one real constant are possible? This question is to better explain my previous question, which got closed. I once asked whether the theory of real addition with a nonzero constant $r$ depends on which $r$ it is. The answer is no. So that proves that there are two types of theories of $(\mathbb{R},+,r)$, depending on whether $r$ is zero or not. Of course the same two-fold classification under multiplication does not work. For example, we can define $r$ to be $1$ by the axiom $(\forall x) x * r = x$, and we can define $r$ to be $-1$ by the conjunction of the negation of $r=1$ and $(\forall x) x * r * r = x$. We can also distinguish between negative and positive $r$ by the axiom $(\exists x)x * x = r$. I conjecture that, in a sense, this is all we can do with just multiplication in the signature. Is this true, or are there in fact more types of theories than just those 5?
 A: Via exponentiation, $(\Bbb R_{>0},\cdot)$ is the same as $(\Bbb R,+)$. Thereafter, $(\Bbb R,\cdot)$ is the same as the disjoint union of $\{0\}$ and $(\Bbb R_{>0},\cdot)$ and the negatives of $(\Bbb R_{>0},\cdot)$; here, the bijecton between positivies comes from multiplication with $-1$, which you saw is definable. Ultimately, everything we can say about $(\Bbb R,\cdot)$ can be expressed in terms of $(\Bbb R,+)$ this way. So indeed, for $(\Bbb R,\cdot,r)$, we have


*

*the same distinctions as fro $(\Bbb R,+,\ln r)$ if $r>0$, namely $r=1$ or $0<r\ne1$.

*the above cases transported to the negatives by multiplication with $-1$, i.e., $r=-1$ or $0>r\ne -1$

*the remaining case $r=0$
So the two cases from addition indeed translate into five cases for multiplication. If there were any additional case for multiplication, this would translate back to an additional case for addition.
A: Yes, you are correct.
We show something stronger: for any $a,b\in\mathbb{R}\setminus\{-1,0,1\}$ with the same sign there is an automorphism $\pi_{a,b}$ of $(\mathbb{R},\cdot)$ with $\pi(a)=b$. A fortiori $(\mathbb{R};\cdot, a)\equiv(\mathbb{R};\cdot,b)$.
Specifically, for $s\in\mathbb{R}\setminus\{0\}$ let $$f_s:\mathbb{R}\rightarrow\mathbb{R}: x\mapsto {\vert x\vert\over x}(\vert x\vert^s).$$ Each such $f_s$ is an automorphism of $(\mathbb{R};\cdot)$, the key points being $u^wv^w=(uv^w)$ and $u^w=v^w\implies u=v$ for $u,v\ge 0$ and $w\not=0$.
Now for $a,b\in\mathbb{R}\setminus\{-1,0,1\}$ with the same sign, taking the appropriate logarithm gives an $f_s$ with $f_s(a)=b$.
