Does the field norm commute with field morphisms? Consider a field extension $K$ of $\Bbb{Q}$, a finite extension $L$ of $\Bbb{Q}_p$ for some fixed prime number $p$, and a field morphism $\sigma \colon K \to L$ such that the diagram
$\require{AMScd}$
\begin{CD}
    K @>\sigma>> L\\
    @AAA @AAA\\
    \Bbb{Q} @>>\iota> \Bbb{Q}_p
\end{CD}
commutes. Is it true that
$$
N_{L/\Bbb{Q}_p}(\sigma(x)) = \iota(N_{K/\Bbb{Q}}(x)) = N_{K/\Bbb{Q}}(x)
$$
for every $x \in K$, where $N_{E/F}$ denotes the field norm of a field extension $E/F$?

Edit: Álvaro Lozano-Robledo showed that the above statement is way too optimistic. Can it be fixed in the following way:
$$
N_{L/\Bbb{Q}_p}(\sigma(x))^{1/d_L} = \iota(N_{K/\Bbb{Q}}(x))^{1/d_K}
$$
where $d_L = [L : \Bbb{Q}_p]$ and $d_K = [K : \Bbb{Q}]$?

No.
Consider $K=\mathbb{Q}(i)$ and let $p\equiv 1 \bmod 4$, so that $i\in \mathbb{Q}_p$. Take $L=\mathbb{Q}_p$, and $\sigma: K\to \mathbb{Q}_p$ an embedding. Let $\alpha=3$. Then, $\iota(N_{K/\mathbb{Q}}(\alpha))=\iota(9)=9$, but $$N_{L/\mathbb{Q}_p}(\sigma(3))=N_{L/\mathbb{Q}_p}(3)=3,$$
because $3\in L=\mathbb{Q}_p$.
