Let $$x$$ be an arbitrary n-dimensional tuple of real coordinates, a real vector.
Let $$\eta$$ be a n×n diagonal matrix where all elements on the diagonal are either +1 or -1.
Let $$A$$ be a totally anti-symmetric matrix such that $$A^{\tiny \textrm{T}} = - A$$. (It follows that A has $$\frac{1}{2} n(n-1)$$ degrees of freedom.)
Let $$q(x_1, x_2, x_3, x_4)$$ be defined as $$\left( x_2 - x_1 \right)^{\tiny \textrm{T}} \, \eta \, \left( x_4 - x_3 \right)$$.
Let
$$\Large x' = X_0 + e^{ \eta A } x $$
be a $$\frac{1}{2} n(n+1)$$-parameter transform of the vector space of x.
Then we see that $$q(x'_1, x'_2, x'_3, x'_4) = q(x_1, x_2, x_3, x_4)$$ holds for any four vectors.
This shows that the property of q is preserved by the transform.
Proof:
First see that $$I = \eta^2$$, $$\eta = \eta^{\tiny \textrm{T}}$$.
Then remember that
$$\sum_{j=0}^n \frac{\left(-1\right)^{j}}{j! \, (n-j)!} = \left{ \begin{array}{lll} 1 & \quad \quad \quad & \textrm{if} \; n = 0 \\ 0 & & \textrm{otherwise} \end{array} \right. $$
Then see that: $$\eta \, \left( \eta A \right)^k = \eta \, \left( \eta A \right)^k \eta^2 = \left( A \eta \right)^k \eta$$ and therefore $$ \left( A \eta \right)^j \, \eta \, \left( \eta A \right)^k = \left( A \eta \right)^{j+k} \eta$$ for all natural numbers j and k.
Then directly calculate:
$$q(x'_1, x'_2, x'_3, x'_4) = \left( x'_2 - x'_1 \right)^{\tiny \textrm{T}} \, \eta \, \left( x'_4 - x'_3 \right)
= \left( (X_0 + e^{ \eta A } x_2) - (X_0 + e^{ \eta A } x_1) \right)^{\tiny \textrm{T}} \, \eta \, \left( (X_0 + e^{ \eta A } x_4) - (X_0 + e^{ \eta A } x_3) \right)
= \left( e^{ \eta A } \left( x_2 - x_1 \right) \right)^{\tiny \textrm{T}} \, \eta \, \left( e^{ \eta A } \left( x_4 - x_3 \right) \right)
= \left( x_2 - x_1 \right) ^{\tiny \textrm{T}} \, \left( e^{ \eta A } \right) ^{\tiny \textrm{T}} \, \eta \, e^{ \eta A } \, \left( x_4 - x_3 \right)
= \left( x_2 - x_1 \right) ^{\tiny \textrm{T}} \, e^{ \left( \eta A \right) ^{\tiny \textrm{T}} } \, \eta \, e^{ \eta A } \, \left( x_4 - x_3 \right)
= \left( x_2 - x_1 \right) ^{\tiny \textrm{T}} \, e^{ A^{\tiny \textrm{T}} \eta^{\tiny \textrm{T}} } \, \eta \, e^{ \eta A } \, \left( x_4 - x_3 \right)
= \left( x_2 - x_1 \right) ^{\tiny \textrm{T}} \, e^{ -1 \, A \eta } \, \eta \, e^{ \eta A } \, \left( x_4 - x_3 \right)
= \left( x_2 - x_1 \right) ^{\tiny \textrm{T}} \, \left( \sum_{j=0}^{\infty} \frac{ (-1)^j \left( A \eta \right)^j}{j!} \right) \, \eta \, \left( \sum_{k=0}^{\infty} \frac{ \left( \eta A \right)^k}{k!} \right) \, \left( x_4 - x_3 \right)
= \left( x_2 - x_1 \right) ^{\tiny \textrm{T}} \, \left( \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \frac{ (-1)^j \left( A \eta \right)^j}{j!} \eta \frac{ \left( \eta A \right)^k}{k!} \right) \, \left( x_4 - x_3 \right)
= \left( x_2 - x_1 \right) ^{\tiny \textrm{T}} \, \left( \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \frac{ (-1)^j \left( A \eta \right)^j \, \eta \, \left( \eta A \right)^k}{j! \, k! } \right) \, \left( x_4 - x_3 \right)
= \left( x_2 - x_1 \right) ^{\tiny \textrm{T}} \, \left( \sum_{n=0}^{\infty} \left( \left( A \eta \right)^n \eta\right) \sum_{j=0}^{n} \frac{ (-1)^j }{j! \, (n-j) ! } \right) \, \left( x_4 - x_3 \right)
= \left( x_2 - x_1 \right) ^{\tiny \textrm{T}} \, \left( \frac{ (-1)^0 }{0! \, 0 ! } \left( \left( A \eta \right)^0 \eta\right) \quad + \quad \sum_{n=1}^{\infty} \left( \left( A \eta \right)^n \eta\right) \sum_{j=0}^{n} \frac{ (-1)^j }{j! \, (n-j) ! } \right) \, \left( x_4 - x_3 \right)
= \left( x_2 - x_1 \right) ^{\tiny \textrm{T}} \, \left( \eta \quad + \quad \sum_{n=1}^{\infty} 0 \right) \, \left( x_4 - x_3 \right)
= \left( x_2 - x_1 \right) ^{\tiny \textrm{T}} \, \eta \, \left( x_4 - x_3 \right) = q(x_1, x_2, x_3, x_4)$$
Q.E.D.
Thus continuous rigid body motions (Translations and Rotations) are isometries of Euclidean geometry with the $$\eta = I$$, while the 10-parameter Poincaré transformations (Translations and spatial Rotations and hyperbolic space-time rotations (Lorentz boots)) are isometries of special relativity with $$\eta = { \tiny \begin{pmatrix} \mp 1 & 0 & 0 & 0 \\ 0 & \pm 1 & 0 & 0 \\ 0 & 0 & \pm 1 & 0 \\ 0 & 0 & 0 & \pm 1 \end{pmatrix} }$$.
