Special relativity and diffeomorphism invariance In studying general relativity (GR) we learn that the Einstein-Hilbert (EH) action $S_{EH}=\int_{M}\mathrm{d}v_{g}R$ (where $\mathrm{d}v_{g}=\mathrm{d}^{4}x\sqrt{-g}$, with $g$ the metric tensor)  is diffeomorphism invariant, i.e. under a diffeomorphsim $\phi^{\ast}:M\to M$ (where $(M,g)$ is the spacetime manifold): $\phi^{\ast}S_{EH}=S_{EH} \iff \mathcal{L}_{\xi}S_{EH}=0$, where $\xi$ is the vector field generating the diffeomorphism. As far as I understand it, one can show this is true as follows: $$\delta_{\xi}S_{EH}=\epsilon\mathcal{L}_{\xi}S_{EH}=\epsilon\int_{M}\big[\mathcal{L}_{\xi}(\mathrm{d}v_{g})R+\mathrm{d}v_{g}\mathcal{L}_{\xi}(R)\big]=\epsilon\int_{M}\big[\mathrm{d}v_{g}\nabla_{\mu}\xi^{\mu}R+\mathrm{d}v_{g}\xi^{\mu}\nabla_{\mu}R\big]\\ =\epsilon\int_{M}\mathrm{d}v_{g}\nabla_{\mu}(\xi^{\mu}R)=0\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;$$ assuming that $\xi$ has compact support, such that it vanishes on the boundary $\partial M$.
This is all well and good, but then how does one show that the action of a theory in the context of special relativity (SR) is not diffeomorphism invariant? Take, for example, electromagnetism (EM), which has the following action $$S_{EM}=-\int\mathrm{d}^{4}x\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$ where we have noted that $\sqrt{-\eta}=1$. Isn't it still the case that $\mathcal{L}_{\xi}(\mathrm{d}v_{\eta})=\mathrm{d}v_{\eta}\partial_{\mu}\xi^{\mu}$ and $\mathcal{L}_{\xi}(F^{\mu\nu}F_{\mu\nu})=\xi^{\mu}\partial_{\mu}(F^{\mu\nu}F_{\mu\nu})$? In which case it would follow that $\phi^{\ast}S_{EM}=S_{EM}$, which would contradict the statement that SR is not diffeomorphism invariant? My understanding of diffeomorphism invariance is that solutions to the Einstein Field equations (EFEs) $G=\text{Ric}(g)-\frac{1}{2}gR(g)=0$ are mapped into solutions, i.e. if $g$ is a solution to the EFEs, related to $g'$ via a diffeomorphism $g=\phi^{\ast}g'$, then $$G(g)=G(\phi^{\ast}g')=(\phi^{\ast}G)(g')=0.$$ However, SR only holds under isometries $\phi^{\ast}\eta=\eta\iff\mathcal{L}_{\xi}\eta =0$. 
I have gotten myself very confused over the issue. Any help on the matter would be very much appreciated.
P.S. I get that so-called active (which we are discussing here) and passive diffeomorphisms are essentially the same in coordinate form (since any infinitesimal diffeomorphism can be expressed in coordinate form as an infinitesimal coordinate transformation), however, the two are very different (at least in a physical sense). There must be a way to understand this without resorting to the coordinate argument?!
 A: The "diffeomorphism invariance" of general relativity is a figure of speech that more properly refers to the invariance of the theory under $\mathrm{GL}(1,n-1)$ gauge transformations that act on all fields like the pushforward under a diffeomorphism but that do not act on the coordinates at all. That is, the gauge transformations of GR are not diffeomorphisms but transformations on the fields that just look like the transformations induced by a diffeomorphism.
The reason GR exhibits this symmetry and other (also diffeomorphism-invariant!) theories do not is that in order for this symmetry to be local you need a (gauge) covariant derivative so that the differentials of the fields transform properly. In GR, the  gauge field for this derivative is provided by the Christoffel symbols, but in theories in which the metric is not dynamical there is no natural candidate for this gauge field.
See also this answer of mine and this answer of mine which also discuss ways in which the "diffeomorphism invariance" of GR is different from the "coordinate change invariance" that all physical theories must enjoy if they are to be consistent.
