It is known that $\Bbb{C}$ is isomorphic as a field to $\Bbb{C}_p$, the completion of $\bar{\Bbb{Q}}_p$ with respect to $\left|\cdot\right|_p$. Clearly, given two such isomorphisms $\varphi_1$ and $\varphi_2$ we can define extensions $\left|\cdot\right|_{p,i} = \left|\varphi_i(\cdot)\right|_p$ of $\left|\cdot\right|_p$ from $\Bbb{Q}$ to $\Bbb{C}$.
Now, I don't expect those two extensions to be equivalent -- especially since I couldn't find any reference for this -- but I've been wondering if the sets $$ U_i := \left\{ x \in \Bbb{C} : \left|x\right|_{p,i} = 1 \right\} $$ coincide.
I'm very new to p-adic analysis, but I think it should be the case. In his book A course in p-adic analysis Alain Robert proves that there is a group isomorphism $$ \begin{align} \psi \colon \Bbb{C}_p^{\times} &\simeq p^{\Bbb{Q}} \times U \\ x = r \cdot \frac{x}{r} &\mapsto (\left|r\right|_p, \frac{x}{r}) \end{align} $$ where $U := \left\{ x \in \Bbb{C}_p : \left|x\right|_p = 1 \right\}$ and $r$ is an element of $p^{\Bbb{Q}} \hookrightarrow \Bbb{C}_p^{\times}$ (after a choice of embedding). In particular the composition $$ \Phi \colon p^{\Bbb{Q}} \times U \xrightarrow{\psi^{-1}} \Bbb{C}_p^{\times} \xrightarrow{\varphi_1^{-1}} \Bbb{C}^{\times} \xrightarrow{\varphi_2} \Bbb{C}_p^{\times} \xrightarrow{\psi} p^{\Bbb{Q}} \times U $$ is a group automorphism. Now suppose that $U_1 \neq U_2$. Then wlog we can find an $x \in U_1$ such that $\left|\varphi_2(x)\right|_p \neq 1$, which means that $\psi(\phi_1(x)) = (1,\phi_1(x))$ and $\Phi(\psi(\phi_1(x))) = (p^q,u)$ with $q \neq 0$. But this is absurd because $\Phi$ restricted to $p^{\Bbb{Q}}$ is a group automorphism.
Remark: Since it wasn't clear, here by "restriction of $\Phi$ to $p^{\Bbb{Q}}$" I actually mean the restriction of $\Phi$ to $p^{\Bbb{Q}} \times \{1\}$ composed with the canonical isomorphism $p^{\Bbb{Q}} \times \{1\} \simeq p^{\Bbb{Q}}$.
Is this argument correct? If so, is there a simpler one?
Here’s your counterexample, with $p=2$. Your element of $\Bbb C$ is $(1+\sqrt{-7})/2$, one chosen root $\alpha$ of $X^2-X+2$. Since this polynomial becomes $X^2+X=X(X+1)$ over $\Bbb F_2$, it splits over the $2$-adic integers $\Bbb Z_2$, by Hensel. That is, in $\Bbb Q_2$, one root is a unit (has $2$-absolute value $1$), the other is a nonunit (has $2$-absolute value less than $1$). Now, any isomorphism from $\Bbb C$ to $\Bbb C_p$ must take $\alpha$ to one of the roots in $\Bbb Q_p$ of $X^2-X+2$. Clearly, there are some that take $\alpha$ to the unit root, some that take $\alpha$ to the nonunit root.