= Solution
If $K$ is complete and $L/K$ is finite, the <unique extension of an absolute value to a finite extension> is
$$
|x|_L=|N_{L/K}(x)|^{1/[L:K]}.
$$
It restricts to the given absolute value because $N_{L/K}(a)=a^{[L:K]}$ for $a\in K$. The usual construction through multiplication by $x$ on the finite-dimensional $K$-vector space $L$, together with completeness, proves that this is the only extending absolute value.
For $\sigma\in\operatorname{Gal}(L/K)$, the map $x\mapsto|\sigma(x)|_L$ is another absolute value extending $|\mathord\cdot|$ on $K$. Uniqueness therefore gives
$$
|\sigma(x)|_L=|x|_L.
$$
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