Lorentz Transformations Vs Coordinate Transformations
A general diffeomorhpism is not an isometry. Or rather, it can be made into an isometry. Consider smooth manifolds $M$ and $N$, with metrics $g$ and $h$. Let $\phi:M\rightarrow N$ be a diffeo. We say that $\phi$ is an isometry if $g=\phi^*h$.
But now, let's forget about $h$. We define it instead as $$ h=(\phi^{-1})^*g. $$
Then $(M,g)$ and $(N,h)$ are automatically isometric as (semi-)Riemannian spaces.
With this said, consider $(M,g)$ to be Minkowski spacetime. Let $\phi:M\rightarrow M$ be a diffeo. Let $X,Y$ be vector fields. Obviously, it is true, that $$ (\phi^{-1})^*g(\phi_*X,\phi_*Y)=g(X,Y), $$ so applying a diffeo to every object on the manifold will preserve relations. But is it true that $$ g(\phi_*X,\phi_*Y)=g(X,Y)? $$ Or alternatively, $$ (\phi^{-1})^*g(X,Y)=g(X,Y)? $$
No. In general it is not true. Those transformations for which $ \phi^*g=g $ in Minkowski spacetime are Poincaré transformations. The (homogenous) linear ones are Lorentz-transformations. This concludes my answer, but here is a (hopefully) illuminating aside.
Although this is in a slightly different context, here is an example, where the difference between isometries or general invertible & structure-preserving transformations make a difference:
Consider local Lorentzian geometry using local (possibly anholonomic) frames. What is the minimum information needed to give the local geometry exactly?
For coordinate frames: The metric components $g_{\mu\nu}$.
For completely general frames: The metric components $g_{ab}$, and the relationship between any one coordinate frame and the general frame, which is $e^\mu_a$ or $\theta^a_\mu$ ($\theta^a=\theta^a_\mu dx^\mu$, $e_a=e^\mu_a\partial_\mu$).
For orthonormal frames: The relationship between any one coordinate frame and the orthonormal frame. Why? Because if $\theta^a_\mu$ is given, then $g_{\mu\nu}=\eta_{ab}\theta^a_\mu\theta^b_\nu$.
So you can see, that despite the fact that all frames are just tools, and they don't have physical/geometric relevance, and thus all frames are equally good, specifying a frame and demanding it to be orthonormal actually gives a metric! There is valuable information content in the fact that a frame is orthonormal, and this info is lost if we use a general frame.
We can of course cast this notion in the language of transformation by noting that given an initial frame, a system of orthonormal frames can be constructed by demaning that two valid frames differ by a Lorentz transformation: $e_{a'}=\Lambda^a_{\ a'}e_a.$ So, despite the fact that any $GL(n,\mathbb{R})$-valued transformation is a good frame transformation, the Lorentz-valued transformations are special. A system of frames for which Lorentz-transforms allowed only specifies a metric uniquely. The associated statement in modern, invariant differential geometry would be that any reduction of the frame bundle $F(M)$'s $GL(n,\mathbb{R})$ into a generalized orthogonal group uniquely yields a semi-Riemannian metric.
You are, I guess, discussing special relativity. In that case, the most natural geometrisation is to postulate that spacetime $\mathcal{S}$ is an real affine space of dimension 4 with a quadratic form of signature $(+,-,-,-)$. Everything follows, among which the two fundamental aspects to study:
- the group $\mathcal{P}$ of affine transforms leaving the quadratic form invariant (called the Poincaré group by physicists);
- there is an infinite family of frames where the quadratic form has a matrix $\mathrm{diag}(+1,-1,-1,-1)$. Any two such frames are mapped onto each others by an element of $\mathcal{P}$.
The first point is about what physicists would call an active transform whereas the second one would be called a passive transform.
You asked about Lorentz transforms: as always for a group of affine transforms, $\mathcal{P}$ is the semi-direct product of the subgroup of translations and of a group $\mathcal{L}$ of linear transforms on the vector space $S$ associated with $\mathcal{S}$. Then $\mathcal{L}$ is called the Lorentz group.
Note, to finish, if it was not totally obvious, that this is totally similar to affine Euclidean geometry and isometries: the only difference is the signature of the quadratic form, which is positive definite, and of course that the dimension is 3 and not 4.