Following the <definition of the indefinite orthogonal group>, we want to show that only the <metric signature> matters.

First we can observe that the exact matrices are different. For example, taking the standard matrix of $O(2)$:
$$
\begin{bmatrix}
1 0
0 1
\end{bmatrix}
$$
and:
$$
\begin{bmatrix}
2 0
0 1
\end{bmatrix}
$$
both have the same <metric signature>. However, we notice that a rotation of 90 degrees, which preserves the first form, does not preserve the second one! E.g. consider the vector $x = (1, 0)$, then $x \cdot x = 1$. But after a rotation of 90 degrees, it becomes $x_2 = (0, 1)$, and now $x_2 \cdot x_2 = 2$! Therefore, we have to search for an <isomorphism> between the two sets of matrices.

For example, consider the <orthogonal group>, which can be defined as shown at <the orthogonal group is the group of all matrices that preserve the dot product> can be defined as:


All indefinite orthogonal groups of matrices of equal metric signature are isomorphic

ID: all-indefinite-orthogonal-groups-of-matrices-of-equal-metric-signature-are-isomorphic