If two Hecke characters cut out the same field, are they Galois conjugates? First question on MathOverflow, I hope it is appropriate for this site. There are two related questions.
Let $K$ be a number field, $G_K = Gal(\overline{K}/K)$, $p$ a prime, and
$$\chi_1,\chi_2:G_K\rightarrow\overline{\mathbb{Q}}_p^\times$$
be continuous characters such that $\ker(\chi_1) = \ker(\chi_2)$. 
My first question is, are the two fields generated by the values of $\chi_1$ and $\chi_2$ the same? That is, does 
$$\mathbb{Q}_p(\chi_1(\sigma), \sigma\in G_K) = \mathbb{Q}_p(\chi_2(\sigma),\sigma\in G_K)$$
As far as I can tell, this is true if $\chi_1,\chi_2$ have finite image, but I'm not sure if it's true if they have infinite image.
My second question is, does there exists $\sigma\in G_{\mathbb{Q}_p} = Gal(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$ such that for all $\tau\in G_K$, $\chi_1(\tau) = \sigma(\chi_2(\tau))$?
If $E = \mathbb{Q}_p(\chi_1(\sigma), \sigma\in G_K) = \mathbb{Q}_p(\chi_2(\sigma),\sigma\in G_K)$, then one could define $\sigma$ by the rule $\sigma(\chi_2(\tau)) = \chi_1(\tau)$, show $\sigma\in Gal(E/\mathbb{Q}_p)$, and take an extension of $\sigma$ to an element of $Gal(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$. Then the answer to the second question would be yes if the answer to the first question is yes.
I don't know how to show that $\sigma$ defined as above is a well defined element of $Gal(E/\mathbb{Q}_p)$.
 A: The answer to both your questions is "no".
Take $K = \mathbf{Q}$. Then there is a unique $\mathbf{Z}_p$-extension of $K$ (contained in $\mathbf{Q}(\zeta_{p^\infty})$) which gives us a surjection $G_K \to \Gamma$ where $\Gamma$ is isomorphic to $\mathbf{Z}_p$.
Now, what are the continuous characters $\mathbf{Z}_p \to \overline{\mathbf{Q}}_p^\times$? It turns out that for every $a \in \overline{\mathbf{Q}}_p^\times$ with $|a - 1| < 1$, there's a unique character sending $1 \in \mathbf{Z}_p$ to $a$. If $a$ is not a root of unity, then this character is an injection.
So pick two values $a, a'$ which aren't roots of unity, don't generate the same extension of $\mathbf{Q}_p$, and aren't Galois-conjugate. Then this gives you two continuous characters $\chi, \chi': G_K \to \overline{\mathbf{Q}}_p^\times$ with the same kernel which are the required counterexamples.
With a little more work you can even find counterexamples with characters that are locally algebraic (and thus arise from algebraic Groessencharacters).
