Can a diffeomorphism make a pseudo-Riemannian manifold more symmetric (so it has more Killing vectors)?

Question

I'm writing a thesis and in one of the papers I'm reading it is said that two spacetimes (Lorentzian manifolds) can be related by a map and one of these spacetimes possesses high symmetry, or it is maximally symmetric (it has 10 Killing vectors in 4 dimensions). It is not explicitly said that it is a diffeomorphism, but what else would it be. The manifolds have the same dimension. I would like to know how is this possible? How come I can find a map $f:M\rightarrow M$ such that the 'new' manifold $f(M)$ has more Killing vectors? It is not that obvious to me. I could imagine a map which spoils the symmetry, so if it is a diffeomorphism then its inverse creates more symmetry. Is this the trick? Let's say that the flow $\Phi_t$ of a vector $\xi$ is not an isometry of $(M,g)$, i.e. $\mathcal{L}_{\xi} g \neq 0$, so with the mentioned $f$, for $(\Phi_t \circ f)$ I should have $\mathcal{L}_{f_{*}\xi} (f^{-1}){}^{*} g=0$.Cool, but I don't see why.