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I am looking at this answer: https://physics.stackexchange.com/a/225417/747. It states:

Let $f\colon U\to V$ be any coordinates transformation on charts of a manifold $U,V\subset\mathcal{M}$ (i. e. a change of coordinates). Under such transformation fields $\phi(x)$ are sent into $\phi'(f(x)) = S(x)\phi(x)$.

In order the equations of motion to be satisfied, one must require certain appropriate conditions on the factor $S(x)$ (in particular one can see that these could be related to the representations of the underlying group of transformations $f$). The set of all allowed operators $S(x)$ defines the symmetry group of the theory for the general mapping $f$ as defined above. In the case of general relativity $f$ are the diffeomorphisms and $S(x)$ span the general linear group (up to isomorphisms and cartesian products).

My question is how far the expression $\phi'(f(x)) = S(x)\phi(x)$ is from general relativity? Can we derive the equations of motion using said expression as a starting point? What steps and additional assumptions to said expression must be made to reach the appropriate equation of motion?

The question I have linked provides a related answer, but not quite: Ideally, I am looking for a step-by-step process starting from $\phi'(f(x)) = S(x)\phi(x)$ as step 1, and the field equations for the final step. If I need additional baggage to get GR, what specifically is the additional baggage, then such baggage should be shown in intermediary steps. Specifically, I work with a system whose solutions is the set of all expressions of the type $\phi'(f(x)) = S(x)\phi(x)$, where $S(x) \in GL(4,R)$. I suspect it is relatable to GR but I do not know how close to it this is. Is a general expression of this type $\phi'(f(x)) = S(x)\phi(x)$ equivalent to GR, enough to derive GR, or merely a solution compatible with GR?

Anon21
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    Does this answer your question? https://physics.stackexchange.com/q/225413/ – Daddy Kropotkin Dec 27 '20 at 14:49
  • @N.Steinle I think the difference is that I am asking for the next step (in more beginner friendly terms): how to get to GR from $\phi'(f(x)) = S(x)\phi(x)$, whereas the linked question is about the relationship between GL(4,R) and Diff(M). – Anon21 Dec 27 '20 at 14:55
  • Then you've answered your own question: "In the case of general relativity f are the diffeomorphisms and S(x) span the general linear group (up to isomorphisms and cartesian products)." So are you really asking for a justification of this statement? – Daddy Kropotkin Dec 27 '20 at 14:59
  • @N.Steinle Yes, but more an explanation in beginner friendlier terms. I have a situation where all I have to work with is an expression of the type $\phi'(f(x))= S(x)\phi(x)$. I suspect it is relatable to GR but I do not know how close to it this is. Can one derive the field equations (or the equations of motion) from said expression? If I need additional baggage to get GR, what specifically is the additional baggage. Ideally, I am looking for a step-by-step process starting from $\phi'(f(x))= S(x)\phi(x)$ as step 1, and the field equations for last step. – Anon21 Dec 27 '20 at 15:07
  • If I were you, I would write that exactly in your question. The clearer you describe your question the more likely you'll get an answer and an answer that is helpful. I do not have an answer though :) Hope my upvote helps – Daddy Kropotkin Dec 27 '20 at 15:30

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The material you've read is just talking about the fact that general relativity can be easily expressed in a coordinate-independent way. However, other theories of physics can also be expressed in a coordinate-independent way. It's just that they weren't originally/traditionally expressed that way. So the answer is no, you can't derive all of general relativity from this.

In this respect, diffeomorphism invariance is different from the kinds of symmetries that we can plug into Noether's theorem. The fact that we can derive conservation laws from those other symmetries means that they are not physically vacuous.

The discussion of GL(4,R) is just a distraction here, and doesn't affect the discussion I've given above. All that's happening is that if you have a field or an infinitesimal displacement, you can use the diffeomorphism to find a more specific local, linear transformation law for these quantities. In the language of finite-dimensional Lie groups, the physically non-vacuous group for relativity is the Lorentz group, which is a subgroup of GL(4,R). Lorentz invariance is the symmetry that is testable (to incredible precision) by experiments.

user283866
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