At least as far as I can tell, historically Galois theory was a more computational tool than it appears now, and https://hsm.stackexchange.com/questions/8099/how-did-the-modern-understanding-of-galois-theory-come-about, How Did Galois Understand the Galois Group? and Intuition behind looking at permutations of the roots in Galois theory seem to say that the "fundamental idea of Galois theory" is to study permutations of the roots, which was first done via symmetric polynomials.
The often presented motivation of Galois theory, the insolubility of the quintic, can also be presented without "modern Galois theory", as in this wonderful video on Arnold's proof: https://www.youtube.com/watch?v=BSHv9Elk1MU&ab_channel=notallwrong. This video also uses the "fundamental idea of Galois theory", i.e. studying permutations of the roots, but this time in the context of basic Riemann surfaces.
My question is this: going back to the late 1800's and early 1900's, how would I realize that the "correct" setting of Galois theory is to study field extensions containing the roots, and automorphisms of fields?
One idea I had was that from one of the links above:
Let $A=\{a_1,...,a_n\}$ be the (distinct) roots of a polynomial $f$ with coefficients in a base field $k$. Then a permutation $\pi$ of the set $A$ is in the Galois group of $f$ (over $k$) if (and only if): for every polynomial $g$ in $R=k[x_1,...x_n]$, $g(a_1,....,a_n)=0 \iff g(\pi(a_1),...,\pi(a_n))=0$.
and hence we are really looking for permutations "that algebra can't see", i.e. the heart of Galois theory is really the observation that there are some permutations of the symbols denoting roots of polynomials that algebra/polynomials are completely unable to distinguish (e.g. algebra can't distinguish between $i$ and $-i$, only our geometric embedding into the plane can). Thus we are not able to permute all the roots willy-nilly ("they are not free")... however, if we linearize and look at the vector space (over the base field $k$) containing all these roots (actually we should look at it as a $k$-algebra because only then can we talk about multiplicative correlations between the roots), then we can try to find a basis for this vector space ("a free set") which allows us to permute willy-nilly. In this perspective, the fact that this $k$-algebra turns out to be a field is sort of irrelevant, I think?