Diffeomorphism invariance in special relativity 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?    
 A: It is indeed "often said that general relativity is diffeomorphism invariant and special relativity is not", which is unfortunate because that's not true. Special relativity is perfectly diffeomorphism invariant, as is Newtonian mechanics and every other well-formulated physical theory. That's because diffeomorphisms (or, more precisely, isometries under which the metric tensor and all other fields on the manifold are appropriately pulled back and pushed forward along the diffeomorphism) simply represent changes of coordinates, and by definition the description of any truly physical quantity is covariant under changes of coordinates.
The only possible sense in which SR is any "less" diffeomorphism invariant than GR is in that Minkowski space happens to admit a natural family of global coordinate systems (Lorentz coordinates) in which the metric happens to take a particularly simple form, which is identically equal to $\pm \mathrm{diag}(-1, 1, 1, 1)$ - so simple that you can completely hide it into the raised and lowered indices, which have a trivial mathematical relation (just flippig the sign of the timelike component). In these coordinates, all of the covariant derivatives reduce to partial derivatives, so for elementary applications it's often easiest to always work in those coordinates and not worry about coordinate transformations. But it's still nothing more than just a particularly convenient choice of coordinates, and you're always free to switch to other coordinates. (Of course, the Riemann curvature tensor will vanish in any system coordinates.) In GR, there's no global choice of coordinates in general, and certainly not one as simple as Lorentz coordinates, so we're forced to deal with all the subtleties of changes of coordinates that we can get away with ignoring in SR.
