Automorphism action on Artin map I would like to refer you to Serge Lang's book on Elliptic Functions $\S10.4$ (page 140).
I'm trying to understand a part in theorem 10 where he proves that
$$\mu(\mathfrak{p})'=\mu(\mathfrak{p}')$$
where $\mu$ is the hecke character of an elliptic curve $A$ over a field $K_0$ which does not contain $\text{End}(E)=k$, $\mathfrak{p}$ is a prime of $K = K_0 k$ above $\mathfrak{p}_0$ which is a prime of $K_0$, and $\;'$ is complex conjugation.
(i)  Let $\theta:k\rightarrow End(A)_{\mathbb{Q}}$ be a normalized isomorphism defined over $K$, $\rho$ be the automorphism of $K=K_0k$ over $K_0$. How do you get  $(*)$ in the following equation?
$$\theta(\mu(\mathfrak{p}))=(\rho\mathfrak{p},K)\overset{(*)}=\rho(\mathfrak{p},K)\rho^{-1}$$
Is this a well known identity? I can't seem to find anything about it in any of the books I have.
(ii) A couple of lines later, he goes on to say

From the formula $\big(\lambda(x)\big)^{\rho}=\lambda^{\rho}(x^{\rho})$, we conclude that $\lambda^{\rho}=\rho(\mathfrak{p},K)\rho^{-1}$ on $K(A^{(l)})$ where $\lambda=\theta(\mu(\mathfrak{p}))$

How did he reach that conclusion?
Is it just because action of $\rho$ on $\theta(\mu(\mathfrak{p}))$ must be $\rho\theta(\mu(\mathfrak{p}))\rho^{-1}$?
But because $\rho\in \text{Aut}(K/K_0)$ and $(\mathfrak{p},K)\in \text{Gal}(K/K_0)$, the action looks a little awkward to me and seems like it should be a composition instead of a conjugation.
 A: (i) For a reference, see Lang's Algebraic Number Theory Chapter I, Section 5 (the top of p. 18). Alternatively, if you simply write out the definition of the Frobenius element, you will see that $(\ast)$ holds.
Explicitly, $(\mathfrak{p}, K/K_0)$ is defined to be an element $\sigma \in \mathrm{Gal}(K/K_0)$ such that $\sigma(\alpha) \equiv \alpha^{N \mathfrak{p}_0} \pmod{\mathfrak{p}}$ for all $\alpha \in \mathcal{O}_K$. Equivalently, $\sigma(\alpha) - \alpha^{N \mathfrak{p}_0} \in \mathfrak{p}$.
Then we have
$$\begin{align*}
(\rho \sigma \rho^{-1})(\alpha) - \alpha^{N \mathfrak{p}_0} &= \rho(\sigma(\rho^{-1}(\alpha))) - \rho((\rho^{-1}(\alpha)^{N \mathfrak{p}_0})\\
&= \rho(\sigma(\beta) - \beta^{N \mathfrak{p}_0})
\end{align*}$$
where $\beta = \rho^{-1}(\alpha) \in \mathcal{O}_K$.
Since $\beta \in \mathcal{O}_K$, we know that $\sigma(\beta) - \beta^{N \mathfrak{p}_0} \in \mathfrak{p}$, hence $\rho(\sigma(\beta) - \beta^{N \mathfrak{p}_0}) \in \rho \mathfrak{p}$. Thus
$$
(\rho \sigma \rho^{-1})(\alpha) \equiv \alpha^{N \mathfrak{p}_0} \pmod{\rho \mathfrak{p}}
$$
for all $\alpha \in \mathcal{O}_K$ and so $(\rho \mathfrak{p}, K/K_0) = \rho (\mathfrak{p}, K/K_0) \rho^{-1}$.
(ii) You already know that $\lambda^\rho = \rho(\mathfrak{p}, K)\rho^{-1}$ on $K$ (this is basically what (i) is), so it is just a matter of showing that the action extends in the natural way when you add the coordinates of the $\ell$-primary torsion. But, from the identity of regular functions $(\lambda(x))^\rho = \lambda^\rho (x^\rho)$, or equivalently, $(\lambda(x^{\rho^{-1}}))^\rho = \lambda^\rho(x)$, if you check what happens to the coordinates under the action you see that $\lambda^\rho$ induces the map $\rho (\mathfrak{p}, K) \rho^{-1}$ on each one. Hence it gives the specified action on all of $K(A^{(\ell)})$.
