If you strengthen the condition $\sigma(\mathbf Q_p) \subset \mathbf Q_p$ to $\sigma$ being the identity on $\mathbf Q_p$, so $\sigma$ is a $\mathbf Q_p$-automorphism of $\mathbf C_p$, then a simple description is possible:

(1) every $\mathbf Q_p$-isomorphism between two finite extensions of $\mathbf Q_p$ extends (in *many* ways) to a field automorphism of $\overline{\mathbf Q_p}$, i.e., to an element of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$,

(2) every element of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ extends uniquely to a continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$,

(3) every continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$ is an isometry of $\mathbf C_p$ and is the continuous extension of a unique element of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$.

Therefore any $\mathbf Q_p$-isomorphism between two finite extensions of $\mathbf Q_p$ extends (in many ways) to a continuous automorphism of $\mathbf C_p$, so in particular for any finite Galois extension $K/\mathbf Q_p$ every element of ${\rm Gal}(K/\mathbf Q_p)$ extends (in many ways) to a continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$. Therefore starting with any non-identity automorphism of a finite Galois extension of $\mathbf Q_p$ we can lift it to a non-identity continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$. For example, the conjugation on $\mathbf Q_p(\sqrt{p})$ extends (in many ways) to a continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$. We only need the first two properties above. The point of the third property is to show which $\mathbf Q_p$-automorphisms of $\mathbf C_p$ come from ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$: the automorphisms that are isometries (hence automatically continuous and thus automatically fixing $\mathbf Q_p$).

Proof of (1): This is a theorem from abstract algebra, having nothing to do with $p$-adic fields: every isomorphism between two fields extends (in many ways) to an isomorphism between their algebraic closures. The proof relies on Zorn's lemma, so it is not constructive. See Lang's *Algebra*, Theorem 2.8 of Chapter V. He proves any embedding $\sigma$ of a field $k$ into an algebraically closed field $L$ extends (by Zorn's lemma) to an embedding of any algebraic extension $E$ of $k$ into the same field $L$, and if $E$ is an algebraic closure of $k$ and $L$ is an algebraic closure of $\sigma(k)$ then he shows that embedding of $E$ into $L$ must be an isomorphism. Apply this theorem with $k$ being a finite extension of $\mathbf Q_p$ and $E = L = \overline{\mathbf Q_p}$.

Proof of (2): Let $\sigma \in {\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$.
Any $\mathbf Q_p$-automorphism of a finite extension of $\mathbf Q_p$ preserves the $p$-adic absolute value on that finite extension, so by viewing two elements of $\overline{\mathbf Q_p}$ inside a finite Galois extension of $\mathbf Q_p$ we get that $\sigma$ is an isometry on $\overline{\mathbf Q_p}$: $|\sigma(\alpha) - \sigma(\beta)|_p = |\alpha - \beta|_p$ for all $\alpha$ and $\beta$ in $\overline{\mathbf Q_p}$. Thus $\sigma$ is uniformly continuous on $\overline{\mathbf Q_p}$, so it extends in exactly one way to a continuous function on $\mathbf C_p$, namely by $\sigma(z) := \lim_{n \rightarrow \infty} \sigma(\alpha_n)$, where $\alpha_n$ is any sequence in $\overline{\mathbf Q_p}$ converging to $z$. It is straightforward to check for all $z$ and $w$ in $\mathbf C_p$ that $|\sigma(z) - \sigma(w)|_p = |z - w|_p$ and then that $\sigma$ is additive and multiplicative on $\mathbf C_p$. To show $\sigma$ on $\mathbf C_p$ is surjective, write $z\in \mathbf C_p$ as a limit of $\alpha_n \in \overline{\mathbf Q_p}$. Since $\sigma$ is an isometry the sequence $\sigma^{-1}(\alpha_n)$ in $\overline{\mathbf Q_p}$ is Cauchy and its limit $w$ in $\mathbf C_p$ satisfies $\sigma(w) = z$.

Proof of (3): Let $\sigma$ be a continuous $\mathbf Q_p$-automorphism of $\mathbf C_p$. Since $\sigma$ fixes each element of $\mathbf Q_p$, we have $\sigma(\overline{\mathbf Q_p}) \subset \overline{\mathbf Q_p}$. The field $\sigma(\overline{\mathbf Q_p})$ is algebraically closed and lies between $\mathbf Q_p$ and $\overline{\mathbf Q_p}$, so $\sigma(\overline{\mathbf Q_p}) = \overline{\mathbf Q_p}$. Thus $\sigma$ restricts to an element of
${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$, so $\sigma$ on $\mathbf C_p$ is the continuous extension of an element of ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$, and it can be the continuous extension of at most one element of the Galois group since $\overline{\mathbf Q_p}$ is dense in $\mathbf C_p$. By the proof of (2), the continuous extension to $\mathbf C_p$ of an automorphism in ${\rm Gal}(\overline{\mathbf Q_p}/\mathbf Q_p)$ is an isometry on $\mathbf C_p$, so $\sigma$ is an isometry on $\mathbf C_p$.

explicitformula? Good luck. $\endgroup$ – KConrad Feb 7 '15 at 3:42anyfield and $C$ isanyalgebraically closed field containing $F$ then any field automorphism of $F$ can be extended to a field automorphism of $C$. (Extensions of homomorphisms of fields to homomorphisms into algebraically closed fields are discussed in Lang'sAlgebra.) So any nontrivial ${\mathbf Q}_p$-automorphism of a finite Galois extension of $\mathbf Q_p$ extends to a nontrivial $\mathbf Q_p$-automorphism of $\mathbf C_p$. $\endgroup$ – KConrad Feb 7 '15 at 4:222more comments