Prove that $f(1)=1$ and $f(a^{-1})=a^{-1}$ if $f:\mathbb Q\to \mathbb Q$ is an automorphism 
Suppose $f:\mathbb Q\to \mathbb Q$ is automorphism of fields and $f(0)=0$, prove that $f(1)=1$ and $f(a^{-1})=a^{-1}$, Hint: look at $f(1*1)$

Attempt:
$f(1)=f(1*1)\overset{\text{Hom'}}{=} f(1)*f(1)$
$1*1$ must be equal to $1$ because $1$ is the unit in the field, then $f(\underbrace{1*1}_{=1})$ therefore $f(1)$ must be equal to $1$ because $f(1)= f(1)*f(1)$
I'm not so sure about my try
 A: The first argument should go into more detail. We know that $f(1) = f(1)^2$. Further, since $f$ is a bijection and $f(0)=0$, it follows that $f(1)\neq 0$. We can then multiply both sides of that first equation by $f(1)^{-1}$ to obtain $1=f(1)$. 
To show that $f(a^{-1})=f(a)^{-1}$, note that $1=f(1) = f(aa^{-1})=f(a)f(a^{-1})$. Multiplying the left and right sides of this equationby $f(a)^{-1}$ gives us $f(a)^{-1}=f(a^{-1})$. 
Now we need to show that $f$ is the identity map. We know that $f(1)=1$, and assuming that $n\in\mathbb{N}$ and $f(n)=n$, then $f(n+1)=f(n)+f(1)=n+1$. It follows by induction that for any $n\in \mathbb{N}$, we have $f(n)=n$. Now assume $n\in \mathbb{N}$. Then $f(-n) = f(-1\cdot n) = f(-1)f(n)=-n$. Note that $f(-1)=-1$ since $f(-1)+1=f(-1)+f(1) = f(-1+1)=f(0)=0$. 
Now let $p/q\in\mathbb Q$, where $p,q\in\mathbb{Z}$ and $q\neq 0$. Then $f(p/q)=f(p)f(q^{-1}) = f(p)f(q)^{-1} = p/q$. We now have that $f=\mathrm{id}_{\mathbb Q}$. 
A: Show that $f(n)=n$ for all natural numbers, then $f(m)=m$ for the integers and finally $f(m/n)=f(m)/f(n)$ for a rational number $m/n$. This implies  $f(r)=r$ for $r=m/n$. More details in the post by florence.
A: Choose $a \in Q$ such that $f(a) \ne 0$. Since $f(a) = f(a.1) = f(a).f(1)$, so $f(1) =1$.
$\forall n \in Z$, we have f(n) = n, so $f(a) = a, \forall a \in Q$. Since $1 =f(1) = f(a.a^{-1})= f(a).f(a^-1)=a.f(a^{-1})$, so $f(a^{-1})= a^{-1}$.
