Suppose space time is the manifold $M $ isomorphic $ \mathbb{R^4}$ whit the metric $-\eta_{00}=\eta_{11}=\eta_{22}=\eta_{33}=1$ in the Cartesian coordinates $\Psi(p)=(x^0,x^1,x^2,x^3)$ for $p \in M $ .
So we have model $A=\bigg \langle {M,\eta,\phi,\frac{\partial}{\partial x^a }} \bigg \rangle $ for general relativity and a model $B=\bigg \langle {M,\eta,\phi,\frac{\partial}{\partial x^a }} \bigg \rangle $
for special relativity where $\phi$ is a scalar field and $\frac{\partial}{\partial x^a }$ vector field.
It is often said that general relativity is diffeomorphism invariant and special relativity not.
For example, suppose we have a diffeomorphism $F:M\rightarrow M$ and the wave
equation $$\eta^{ab}\frac{\partial \phi}{\partial x^a }\frac{\partial \phi}{\partial x^b }=0$$
We have by pullback $$F^*\left(\frac{\partial }{\partial x^a }\right)=\frac{\partial G^{b}}{\partial x^a }\frac{\partial}{\partial x^b },$$ $$F^*\phi=\phi \circ F$$ and $$F^*(\eta_{ab})=\frac{\partial F^{k}}{\partial x^a }\frac{\partial F^{j}}{\partial x^b }\eta_{kj}$$ where $G=F^{-1}$
So $$F^*\left(\eta^{ab}\frac{\partial \phi }{\partial x^a }\frac{\partial \phi}{\partial x^b }\right)=\eta^{ab}\frac{\partial (\phi \circ F)}{\partial x^a }\frac{\partial (\phi \circ F)}{\partial x^b}$$
that is $\phi \circ F$ is a solution if is a solution $\phi$.
Is the reason that this does not work in special relativity is that in special relativity the metric $\eta$ has to be fix. That is $$F^*B=\bigg \langle {M,F^*\eta,F^*\phi,F^*\phi\frac{\partial}{\partial x^a }} \bigg \rangle $$ is not a model of special relativity because $ F^*\eta \neq \eta$, and in general relativity if $A$ is model than $C=\bigg \langle {M,g,\phi,\frac{\partial}{\partial x^a }} \bigg \rangle $ is a different model than $A$, but a model of general relativity.
What i mean is,though $F^*B$ and $B$ have equivalent solution they are not equivalent models because $F^*B$ is not a model of special relativity.
Is this the reason why special relativity is not diffeomorphism invariant?
