I found this question which discusses that all orthogonal matrices are rotations/reflections, since the map $X\rightarrow AX$ preserves the scalar product, with $A$ an orthogonal matrix (see proof in the question mentioned).
Before I had found a resource that said that a $4$ (or higher) dimensional rotation matrix should be of the form: $$\begin{bmatrix} \cos(\theta) & -\sin(\theta) & 0 & 0 \\ \sin(\theta) & \cos(\theta) & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{bmatrix}$$
Basically, it has the usual $\cos,\sin$ structure of the $2$ and $3$D rotations, with the off-diagonal elements being $0$ and the diagonal elements being $1$.
This definition of rotation considers that operation as going from one axis to another, i.e. rotation in $2$D is around a point, where the $x$-axis goes towards the $y$-axis, rotation in $3$D is around a line and this time there are different options: $x$-axis goes towards $y$-axis or $z$-axis, $y$-axis towards $z$-axis, etc. This suggests that rotation in $4$D happens over a plane, etc. This constructions imposes a certain structure on rotation matrices.
This basically implies that there orthogonal matrices that are not rotation in the sense I just defined.
So, is it correct to say/more precise to say that all orthogonal matrices are rotations/reflections in that the map $X\rightarrow AX$ preserves the scalar product. But if one considers the definition I used of rotation, then not all orthogonal matrices are rotation matrices? Or is there a way to bridge these two seemingly different definitions?
I vaguely remember when I was reading about this that they mentioned through a change of basis you reach any orthogonal matrix from a rotation matrix with the structure I mentioned above. Is this true?