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.
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 $\epsilon_p=\pm1$.
$$
\mathbb{Q}^*\cong gr(-1)\times\prod_{p\in P} gr(p).
$$
Here, P is the set of primes.
From this it follows that
$$
\prod_{p\in P} gr(-1)&lt;Aut(\mathbb{Q}^*).
$$