Are $\sigma_1(x), \dots, \sigma_n(x)$ linearly independent for some $x \in K \setminus F$? Today I tried to prove exactly what this question asks. In the notation of the question, I tried to show that if $G = \{\sigma_1, \dots, \sigma_n\}$ and $x \in K \setminus F,$ then $\{\sigma_1(x), \dots, \sigma_n(x)\}$ is a basis. After being unable to prove linear independence for 10 minutes, I realized that I already came across the statement the 1st time I took a field theory course, the proof involved several lemmas, and there was no simple path. Thus, I decided to look up the proof and stumbled across the linked question, which as it turns out is roughly how I learned it.
Upon typing this question, I realized that my approach can't possibly work anyways since $\mathbb{C} = \mathbb{R}(i)$ but $\{i, -i\}$ isn't a basis over $\mathbb{R}.$
One possible question remains: does there always exist some $x$ such that $\sigma_1(x), \dots, \sigma_n(x)$ is a basis? For the previous example, we can take $x = 1+i.$ In general, I'm not sure how to proceed.
 A: Everything that you ask for is mentioned in this blurb by K. Conrad. I give a brief summary.
What you're asking for is a normal basis.
While $\{\iota, -\iota\}$ is not an $\Bbb R$-basis of $\Bbb C$, the tuples $(1, 1), (\iota, -\iota)$ do form a $\Bbb C$-basis of $\Bbb C^2$.
More generally:

Theorem 1. Let $e_1, \ldots, e_n$ be an $F$-basis of $K$. Then, the $n$ $n$-tuples
$$(\sigma_1(e_i), \ldots, \sigma_n(e_i))$$
form a $K$-basis of $K^n$.

Using the above, one can prove that every finite Galois extension of an infinite field has a normal basis. The idea is that if the set $\{\sigma_1(x), \ldots, \sigma_n(x)\}$ is independent if the $n \times n$ matrix $[\sigma_i^{-1} \sigma_j (x)]_{i, j}$ is invertible.
To show that this is invertible for some $x$, one constructs a multivariate polynomial $f(X_1, \ldots, X_n) \in K[x_1, \ldots, x_n]$ as a certain determinant. Then, using linear independence of characters and Theorem 1 above, one shows that $f$ is not the zero polynomial (by showing that it is nonzero at some point in $K^n$). But now, since $F$ is infinite, one can find a non-solution $(b_1, \ldots, b_n) \in F^n$. Then, $x = \sum b_i e_i$ does the trick.
For the finite case, one uses the fact that all finite extensions of finite fields are cyclic. Thus, the Galois group looks like $\langle \sigma \rangle$. Then, one considers $K$ as an $F[X]$ module with action given via $\sigma$. Linear independence of characters forces that $\operatorname{ann}_{F[X]}(K) = \langle X^n - 1 \rangle$. The structure theorem for modules over a PID now implies the existence of a single element $a_0 \in K$ such that $\operatorname{ann}_{F[X]}(a_0) = \langle X^n - 1 \rangle$.
This induces a $K$-linear inclusion $K[X]/\langle X^n - 1\rangle \hookrightarrow L$ given by $[X] \mapsto a_0$.
A dimension check shows that this map is a surjection and hence, the image of the basis $\{1, [X], \ldots, [X^{n - 1}]\}$ must be a basis. But this image is precisely $\{a_0, \sigma(a_0), \ldots, \sigma^{n - 1}(a_0)\}$.
