Why must an automorphism of an extension of $\mathbb{Q}$ send 1 to a rational number? This is from a video I was watching that claimed this:

If $\phi$ is an automorphism of an extension field $F$ of $\mathbb{Q}$, then $\phi(q)=q$ for all $q\in\mathbb{Q}$.

The proof started by assuming that $\phi(1)=q$ for some $q\in\mathbb{Q}$. I understand the stuff that followed. But the explanation never said why we don't have to consider the case in which $\phi$ sends $1$ to something in $F\setminus\mathbb{Q}$. 
How come we know an automorphism of an extension of $\mathbb{Q}$ must send $1$ to something in $\mathbb{Q}$ to begin with?
(Thanks in advance.)
 A: Let 
$\phi:F \to F \tag 1$
be any field automorphism of the field
$F \supset \Bbb Q; \tag 2$
then
$(\phi(1))^2 = \phi(1) \phi(1) = \phi(1^2) = \phi(1), \tag 3$
which implies
$\phi(1) \in \{0, 1 \}; \tag 4$
now
$\phi(1) = 0 \Longrightarrow \phi(c) = \phi(c \cdot 1) = \phi(c)\phi(1) = \phi(c) \cdot 0 = 0, \tag 5$
which we rule out since $\phi$ is a field automorphism; thus we see just why
$\phi(1) = 1; \tag 6$
it follows that
$n \in \Bbb N \Longrightarrow \phi(n) = \phi(\underbrace{1 + 1 + \ldots + 1}_{\text{ n times}}) = \underbrace{\phi(1) + \phi(1) + \ldots + \phi(1)}_{\text{n times}} = n; \tag 7$
and also for $n \in N$,
$n + (-n) = 0 \Longrightarrow \phi(n) + \phi(-n) = 0 \Longrightarrow \phi(-n) = -\phi(n), \tag 8$
since in addition $\phi(0) = 0$ (always!), it follows from (7) and (8) that
$z \in \Bbb Z \Longrightarrow \phi(z) = z; \tag 9$
thus, for
$\dfrac{r}{s} \in \Bbb Q, \; r, s \in \Bbb Z, \tag{10}$
$s \phi \left (\dfrac{r}{s} \right ) = \phi(s)\phi \left (\dfrac{r}{s} \right ) = \phi \left ( s\dfrac{r}{s} \right ) = \phi(r) = r, \tag{11}$ 
whence
$\phi \left (\dfrac{r}{s} \right ) = \dfrac{r}{s}, \tag{12}$
and we have shown that
$\forall q \in \Bbb Q, \; \phi(q) = q, \tag{13}$
as per request.  $OE\Delta$.
A: I have no idea why specifically $q\in \Bbb Q$. However, we know any homomorphism sends $1$ to an idempotent element. In a field there are only two idempotent elements. So $\phi(1)=1$ is quite easy to prove. And from there $\phi(q)=q$ for any $q\in \Bbb Q$ follows.
