Any field automorphism of $p$-adics is continuous? How do I see that any field automorphism of $\mathbb{Q}_p$ is continuous?
 A: There is only one field automorphism of $Q_p$, which is maybe your motivation for asking this question.
The idea is to show that the valuation does not change. Suppose that $\phi$ is your field automorphism, and that a nonzero element $a$ of $Q_p$ is decomposed as $a = p^n u$, $u$ is a $p$-adic unit. Then $\phi(a) = p^n \phi(u)$. Hence if we show that $\phi(u)$ is a p-adic unit (a unit in $Z_p$), then it follows that $\phi(a)$ and $a$ have the same $p$-adic valuation. It follows from this that $\phi$ is continuous.
To do this, one wants an intrinsic algebraic characterization of the $p$-adic units. Here it is:
$x \in Q_p^*$ is a p-adic unit iff $x^{p-1}$ has $n$th roots for infinitely many values of $n$.
Proof: We are trying to solve the equation $y^n = x^{p-1}$ inside of $Q_p$ for some $n$. 
Solving $p$-adic equations usually comes down to Hensel's lemma (the formal inverse function theorem), so we just need to solve this equation mod p and check that the derivative is nonzero at the solution.
If $x$ is a unit, this is the claim that there are infinitely many $n$ coprime to $p$ for which $y^n = 1$ - there is obviously a solution in $Z/pZ$, and then we use coprimality of $n$ to $p$ to lift by Hensels lemma.
On the other hand, if there are infinitely many $n$ for which this equation can be solved, we just compute valuations to get $n v(y) = (p -1) v(x)$. Now if $v(x) \not = 0$ we deduce that $n$ divides $(p-1)$ for infinitely many $n$, which is a contradiction. Therefore $v(x) = 0$, so it is a unit.
Reference: https://books.google.com/books?id=cDXoBwAAQBAJ&pg=PA53&lpg=PA53&dq=automorphisms+of+QP&source=bl&ots=fCvzB_rvgx&sig=EZ1YsOSZZ6YtfkICw0WV9gkzUlI&hl=en&sa=X&ved=0ahUKEwj3w5aLouPJAhVGPz4KHUUAB3MQ6AEIHTAA#v=onepage&q=automorphisms%20of%20QP&f=false
A: HINT:
The automorphism is the identity on $\mathbb{Q}$. Let us show that $v_p(\phi(x)) = v_p(x)$ for all $x \in \mathbb{Q}_p$. Assume otherwise. Then multiplying by some power of $p$ (and raising to some power)  we may assume that $v_p(x) \ge 3$ while $v_p(\phi(x)) = 0$. Let $y \in \mathbb{Z}$ so that $1 + y \phi(x)$ is not a square mod $p$. Then $1 + y \phi(x)$ is also not a square in $\mathbb{Q}_p$. However, $1 + y x$ is a square in $\mathbb{Q}_p$ and $\phi(1 + y x) = 1 + y \phi(x)$, contradiction. 
It turns out that $\phi$ will be the identity, since $\mathbb{Q}$ is dense in $\mathbb{Q}_p$.
