Automorphism from $\Bbb Q$ to $\Bbb Q$ Any automorphism of the group $\Bbb Q$ under addition is of the form $x\to qx$ for some $q\in \Bbb Q$. 
I don't know how to proceed in this. Even if i say that $1\to q$, I can't claim that $x\to qx$ since $\Bbb Q$ is not cyclic.  
 A: If $1 \to q$, then


*

*$2 \to ?$

*$1/2 \to ?$

A: The following answer is given in the spirit of overly-pedantic exposition, spelling out ever detail for novice student.  This is not the kind of presentation that I would generally encourage in written form, but I might go into this much detail in a lecture to an introductory algebra class.


Theorem: Every automorphism of the additive group $\mathbb{Q}$ is of the form $x \mapsto qx$ for some $q \in \mathbb{Q}$.

Proof:  Let $\varphi$ be an automorphism of the additive group $\mathbb{Q}$.  That is, suppose that $\varphi: \mathbb{Q} \to \mathbb{Q}$ has the following properties:

*

*the map $\varphi$ is bijective (that is, it is injective and surjective), and

*if $x,y\in \mathbb{Q}$, then $\varphi(x+y) = \varphi(x) + \varphi(y)$.

To show that $\varphi(x) = qx$ for some $q \in \mathbb{Q}$, we can "bootstrap" our way up.  First, we need to know where $1$ goes.  The two properties given above don't put any restriction on the image of $1$, so all we know is that there is some rational number $q$ such that $\varphi(1) = q$.  Fortunately, that's good enough.
Next, an induction argument can be used to show that if $m\in\mathbb{N}$, then $\varphi(m) = qm$.  The base case ($m=1$) is straight-forward:  $\varphi(1) = q = 1\cdot q$.  Now suppose that $\varphi(k) = qk$ for some natural number $k$.  Then
\begin{align}
\varphi(k+1) &= \varphi(k) + \varphi(1) && (\text{property (2), above}) \\
&= kq + q && (\text{induction hypothesis}) \\
&= (k+1) q,
\end{align}
which completes the induction argument.
Any homomorphism must send the identity to itself, and so for any $m\in\mathbb{N}$
$$0 = \varphi(0) = \varphi(m-m) = \varphi(m) + \varphi(-m) = qm + \varphi(-m), $$
which implies that $\varphi(-m) = -qm$.  Thus $\varphi(m) = qm$ for any integer $m$.
Another quick induction argument can be used to show that if $r \in \mathbb{Q}$ is arbitrary and $n$ is any natural number, then
$ \varphi(nr) = n \varphi(r). $
To finish this part of the argument, let $x = m/n \in \mathbb{Q}$ be arbitrary.  Then
$$ n \varphi(x) = \varphi(nx) = \varphi\left( n \frac{m}{n} \right) = \varphi(m) = qm \implies \varphi(x) = \frac{qm}{n} = qx. $$
That is, if $x \in \mathbb{Q}$, then $\varphi(x) = qx$, which is the desired result.

It might be worth noting that the proof above shows that if $\varphi: \mathbb{Q} \to \mathbb{Q}$ is any homomorphism, then there must be some $q\in\mathbb{Q}$ such that $\varphi(x) = qx$.  Since every automorphism is a homomorphism, the theorem is proved.  However, this argument does not immediately tell us that any automorphisms exist, nor does it imply that every map of the form $x \mapsto qx$ is an automorphism.  For this, it is necessary that such maps are bijective.
Before doing this, it is, perhaps, worth noting that $x \mapsto 0$ is a homomorphism of the additive group $\mathbb{Q}$ of the form $x \mapsto qx$ (here, $q=0$), but this map is not an automorphism (for obvious(?) reasons).  So we don't have to worry about zero—this doesn't give us an automorphism.
On the other hand, let $q$ be a nonzero rational number, and let $\varphi(x) = qx$.  To see that $\varphi$ is injective, fix $x,y \in \mathbb{Q}$ and suppose that $\varphi(x) = \varphi(y)$.  Then
$$ \varphi(x) = \varphi(y) \iff qx = qy \iff x = y. $$
Thus $\varphi$ is injective.
Now suppose that $x \in \mathbb{Q}$.  Then
$$ x = q \frac{x}{q} = \varphi\left( \frac{x}{q} \right), $$
where $x/q$ is a rational number.  Thus $\varphi(x)$ is surjective.
This proves a slightly stronger statement that the original:

Theorem: A map $\varphi$ from the additive group $\mathbb{Q}$ to itself is an automorphism if and only if $\varphi(x) = qx$ for some $q \in \mathbb{Q} \setminus \{0\}$.

A: Let $f:\mathbb{Q}\to \mathbb{Q}$ be such a homomorphism, and let $q = f(1)$, then


*

*For any $m\in \mathbb{Z}$,
$$
f(m) = mf(1) = mq
$$

*For any $r = m/n \in \mathbb{Q}$, you have
$$
nf(r) = f(m) = mq \Rightarrow f(r) = q(m/n) = qr
$$

