On cardinality of the two groups $\operatorname {Aut} \mathbb Q$ and $\operatorname {Aut} \mathbb R .$ $\mathbf {The \ Problem \ is}:$ What is the cardinality of $\operatorname {Aut} \mathbb Q$ and $\operatorname {Aut} \mathbb R$  (where $\mathbb R$ and $\mathbb Q$ are groups under usual addition) ???
$\mathbf {My \ approach} :$ Actually, I thought this problem during trying that if two groups $G_1$ and $G_2$ are isomorphic, then $\operatorname {Aut} G_1$ and $\operatorname {Aut} G_2$ are too whether $G_1$ and $G_2$ are finitely generated or not . 
Here, for each $n \in \mathbb N$ , define : $f_n : r \mapsto nr$ for each rational $r$ ; and $f_\alpha : x \mapsto \alpha x$ for each real $x$ ; then $f_n \in \operatorname {Aut} \mathbb Q$ and  $f_\alpha \in \operatorname {Aut} \mathbb R .$
Now, I can't approach further about finding other automorphisms, though there are at least $\mathbb R$- many permutations of $\mathbb Q$, all of them are obviously not automorphisms.
 A: There cannot be more automorphisms of $(\Bbb R,+)$ than there are permutatiosn of $\Bbb R$.
On the other hand,
$\Bbb R$ is a $|\Bbb R|$-dimensional $\Bbb Q$-vector space. If we split a basis into two disjoint subset $B_1,B_2$ of same cardinality, then any bijection $B_1\to B_2$ gives rise to an (involutory) automorphism of $(\Bbb R,+)$, and the number of such bijections is the same as the number of  permutations of $\Bbb R$. So the answer for $(\Bbb R,+)$ is 
$$|\Bbb R|^{|\Bbb R|}$$
A: This is not a complete answer (but too long to be a comment). I don't know about the cardinality of $Aut(\mathbb{R})$.
Note that if $\varphi$ is an automorphism of $(\mathbb{Q}, +)$, then we have
$$ \varphi(1) = \varphi (\sum_{i=1}^n \frac{1}{n}) = \sum_{i=1}^n \varphi(\frac{1}{n})= n \varphi(\frac{1}{n}) $$
Thus, we have
$$ \varphi(\frac{1}{n}) =\frac{\varphi(1)}{n}$$
Using a similar trick as before we get
$$ \varphi(\frac{m}{n}) = \frac{m}{n}  \varphi(1).$$
This means that all automorphisms of $(\mathbb{Q}, +)$ are of the form
$$ x \mapsto q x $$
for some $q\in \mathbb{Q}$.
