Homomorphisms from $(\Bbb Q,+)$ to $(\Bbb Q,+)$ and from $(\Bbb Q^*,\cdot)$ to $(\Bbb Q^*,\cdot)$ Hope this isn't a duplicate.
Considering homomorphisms from $(\Bbb Q,+)$ to itself, I notice that all the homomorphisms are of the form $\phi(x) = ax$ , where $a=\phi(1)$ . So there are exactly $\aleph_0$ number of homomorphisms from $(\Bbb Q,+)$ to itself and except the trivial 0-homomorphism, all others are onto-homomorphism . Is it correct?
On the other hand, considering homomorphisms from $(\Bbb Q^*,\cdot)$ to itself, I get that $\phi(1) =1$ and also get some additional restrictions like if $x \in \Bbb Q^*$ and $x$ is a square number then $\phi(x) \neq y$ , where y is either (negative) or (positive but does not admit a rational square root) . Are there some other restrictions on a general homomorphism  from $(\Bbb Q^*,\cdot)$ to itself ? 
The following maps,(i) $ x \mapsto x$ and (ii) $x \mapsto x^{-1}$ (since, $(\Bbb Q^*,\cdot)$ is abelian) are automorphisms of $(\Bbb Q^*,\cdot)$ . How can one find the complete list of $Aut(\Bbb Q^*,\cdot)$ ?
Thanks in advance for help.
 A: There are lot more automorphisms of $\Bbb Q^*$. By the fundamental theorem
of arithmetic, $\Bbb Q^*\cong (\Bbb Z/ 2\Bbb Z)\times\bigoplus_p\Bbb Z$
where the $p$ range over all primes. The automorphism group of $\Bbb Q^*$ 
is uncountable, as for instance we can consider the maps fixing $-1$
and taking $p$ to $\epsilon_p p$ with $\epsilon_p=\pm1$.
A: The endomorphism ring of $(\mathbb{Q},+)$ is the field $\mathbb{Q}$, the isomorphism being $f\mapsto f(1)$, for $f$ an endomorphism of $(\mathbb{Q},+)$.
The group $(\mathbb{Q}^*,\cdot)$ is isomorphic to $C_2\times\mathbb{F}$, where $F$ is a free abelian group on a countable basis and $C_2$ is the two-element group (I'll write both additively). If we represent elements of this group as columns $\left[\begin{smallmatrix}a\\x\end{smallmatrix}\right]$, for $a\in C_2$ and $x\in F$, endomorphisms can be represented as matrices
$$
\begin{bmatrix}
\alpha & \beta \\
0 & \gamma
\end{bmatrix}
$$
where $\alpha$ is an endomorphism of $C_2$, $\beta\colon F\to C_2$ and $\gamma$ is an endomorphism of $F$. The composition of maps can be written as formal matrix multiplication, so in order to have an automorphism, we need $\alpha$ to be invertible (hence be the identity on $C_2$) and $\gamma$ to be an automorphism of $F$; there's no restriction on $\beta$. Then
$$
\begin{bmatrix}
1 & \beta \\
0 & \gamma
\end{bmatrix}^{-1}=
\begin{bmatrix}
1 & -\beta\gamma^{-1} \\
0 & \gamma^{-1}
\end{bmatrix}
$$
Thus you can get the cardinality of the set of all automorphisms:


*

*it is at least the size of the automorphism group of $F$, which is $2^{\aleph_0}$, because it contains all automorphisms defined by permutations of the basis; 

*it can't be greater than $2^{\aleph_0}$, because the cardinality of all maps $\mathbb{Q}^*$ to itself is $2^{\aleph_0}$.
