Automorphisms of $\mathbb{R^*}$ as a group I was just thinking randomly about groups and cardinality of sets then I thought of the problem

What will be the cardinality of the group of all automorphisms of the multiplicative group $\mathbb{R^*}$?

My Attempt: At first I thought that there's only the identity automorphism and I also got some knowledge about it from few questions over here but then I realised that the questions here was about Ring automorphisms mainly.
Then I thought if $f(x) = x^{2n+1}$ where $n \in \mathbb{N}$. Then $f$ is automorphism.
So I could say that $Aut(\mathbb{R^*})$ is at least countably Infinite but then I couldn't go any further.
Can anyone give me some ideas about it?
I'd prefer if it involves simple group theory approaches.
It's okay if it's not being possible to do so.
Edit : I was also looking for functions $f$ satisfying $f(xy)=f(x)f(y)$. But as I can't say anything about $f$ rather than it's bijection and preserves group structure, I could not find any form of $f$.
 A: The following expands the comments above into an answer:
First, note that the structure $(\mathbb{R}_{>0};\times)$ is isomorphic to the structure $(\mathbb{R};+)$ (think e.g. about the map $x\mapsto ln(x)$). The latter is a bit easier to think about. In particular, if you consider it as a vector space over $\mathbb{Q}$ it's easy to see that its dimension is $2^{\aleph_0}$; any permutation of a basis extends to an automorphism of the whole, and so $Aut(\mathbb{R};+)$ has cardinality at least $2^{2^{\aleph_0}}$. Since that's also an obvious upper bound (that's the number of functions from $\mathbb{R}$ to $\mathbb{R}$), we get $$\vert Aut(\mathbb{R}_{>0};\times)\vert=2^{2^{\aleph_0}}.$$
What about $(\mathbb{R};\times)$? It turns out that there is no real difference between $(\mathbb{R}_{>0};\times)$ and $(\mathbb{R};\times)$  as far as automorphisms are concerned:

Every automorphism of $(\mathbb{R}_{>0};\times)$ extends uniquely to an automorphism of $(\mathbb{R};\times)$, and every automorphism of $(\mathbb{R};\times)$ restricts to a (trivially) unique automorphism of $(\mathbb{R}_{>0};\times)$.

The key point is the following: a real number $r$ is non-negative iff it has a square root. Since this is a purely multiplicative property, it can't be affected by automorphisms - every automorphism of $(\mathbb{R};\times)$ must send positive reals to positive reals and negative reals to negative reals. Thinking a bit more about this gives the result above.
This gives the desired cardinality result: $$\vert Aut(\mathbb{R};\times)\vert=2^{2^{\aleph_0}}.$$
