Automorphism group of ($\mathbb{R}^{\times}, \cdot$) Let ($\mathbb{R}^{\times}, \cdot$) be the multiplicative group of non-zero real numbers. I want to find the automorphism group of ($\mathbb{R}^{\times}, \cdot$). My guess is that it is the group of all rational numbers $p/q$ such that $p, q$ are odd integers which are co-prime. 
We know that any automorphism of the additive group of real numbers ($\mathbb{R}, +$) is of the form $x \mapsto \lambda x$. Also, ($\mathbb{R}_{>0},\cdot $) is isomorphic to ($\mathbb{R}, +$), where ($\mathbb{R}_{>0},\cdot $) denotes the multiplicative group of positive reals.
$\require{AMScd}$
\begin{CD}
(\mathbb{R}_{>0}, \cdot) @>{\phi}>> (\mathbb{R}_{>0}, \cdot)\\
@V{\log(x)}VV  @VV{\log(x)}V\\
(\mathbb{R},+) @>{\lambda x}>> (\mathbb{R}, +)
\end{CD}
By the diagram above, any automorphism $\phi$ of $(\mathbb{R}_{>0}, \cdot)$ is of the form $$ \phi(x) = \exp(\lambda \log(x)) = x^\lambda$$where $\lambda \in \mathbb{R}^{\times}$. As any automorphism $\psi$ of ($\mathbb{R}^{\times}, \cdot$) should take positive numbers to positive numbers, $\psi$ must restrict to an automorphism on ($\mathbb{R}_{>0},\cdot $). The only automorphisms of ($\mathbb{R}_{>0},\cdot $) which extend to the automorphisms of ($\mathbb{R}^{\times}, \cdot$) are $x \mapsto x^\lambda$ such that $\lambda = \frac{p}{q}$ such that $p,q$ are odd integers which are co-prime. This is because: 


*

*For irrational $\lambda$'s, the map is not well defined for negative reals.    

*For $\lambda = \frac{p}{q}$, and $q$ even, the map is not well defined for negative reals.

*For $\lambda = \frac{p}{q}$, and $p$ even, the inverse of this map does not exist.


From here on, I don't know how to proceed in proving my guess, if it is right in the first place.
 A: We have that $(\mathbb{R}^*,\cdot)\simeq \{-1,1\}\times (\mathbb{R}^+,\cdot)$. 
For any automorphism $f:(\mathbb{R}_{>0},\cdot)\to(\mathbb{R}_{>0},\cdot)$ we can define
$$f'(\epsilon,x) = (\epsilon,f(x))$$
and this is an automorphism of $(\mathbb{R}^*,\cdot)$. Does this define all automorphisms? Suppose conversely that
$$g(\epsilon,x) = (g_1(\epsilon,x),g_2(\epsilon,x))$$
is an automorphism. If we compose with the absolute value, we obtain $|g|:(\mathbb{R}^*,\cdot)\to (\mathbb{R}^+,\cdot)$ given by
$$|g|(\epsilon,x) = g_2(\epsilon,x)$$
The kernel is exactly $\{g^{-1}(-1,1), 1\}$, and since $(-1,1)$ is the only element of order $2$, $g(-1,1) = (-1,1)$, hence $g_2(\epsilon,x)$ is independent of $\epsilon$ and hence arises from an automorphism $g_2:(\mathbb{R}_{>0},\cdot)\to (\mathbb{R}_{>0},\cdot)$. Thus
$$g(\epsilon,x) = (g_1(\epsilon,x),g_2(x))$$
If $g_1(\epsilon,x) = (-1,g_2(x))$, then $g(\epsilon,x) = (-1,1)g(-\epsilon,x)$. Since $g(\epsilon,x)^2 = g(-\epsilon,x^2) = g(-\epsilon,x)^2$, it follows that $-\epsilon = 1$. Hence
$$g(\epsilon,x) = (\epsilon,g_2(x))$$
This characterizes the automorphisms as being automorphisms of $(\mathbb{R}_{>0},\cdot)$ linearly extended to be able to pull out minus signs. These will usually no longer be given by a single formula involving exponents, even if they are continuous. The continuous automorphisms end up being
$$f(r) = \begin{cases}
r^a&\mbox{ if }r>0\\
-r^a&\mbox{ if }r<0
\end{cases}$$
