Consider theorem 1.7 from chapter III of 'Elementary differential geometry' by O'Neill. It says that:

Theorem 1.7: If $\phi$ is an isometry of $E^3 $, then there exists a unique translation $T$ and a unique linear transformation $\psi$ such that $\phi = T\psi$.

Now, consider the relativistic interval defined on $R^n$ as usual, i.e. $I(x,y) = \sqrt{\sum (x_i - y_i)^2 - c^2(x_n - y_n)^2}$. We say that $\phi$ is a minkowskian isometry from $R^n$ to $R^n$ just in case for any x,y $\in R^n$, $I(x,y) = I(\phi(x), \phi(y))$.

The question is: does the theorem 1.7 transfer immediately to the minkowskian isometry? That is, can we decompose uniquely the Minkowskian isometry into a translation $T'$ in Minkowski space and a linear transformation $\psi'$ (which would be represented by a Lorentz matrix) ?