I've seen "hyperbolic rotation" - from this: generalization to multisection rotation: is this possible? This question is more in recreational mathematics area 
By accident I came across the concept of "hyperbolic rotation" where we use a matrix containing $\cosh$ and $\sinh$ instead of the trigonometric $\cos$ and $\sin$ for the rotation by matrix-multiplication such that we have:
$$ \tag{trigonometric} \qquad T_t(\varphi) = \begin{bmatrix} \cos \varphi &  \sin \varphi \\  -\sin \varphi &  \cos \varphi  \end{bmatrix} $$
$$ \tag{hyperbolic} \qquad T_h(\varphi) = \begin{bmatrix} \cosh \varphi &  \sinh \varphi \\  \sinh \varphi &  \cosh \varphi  \end{bmatrix} $$
A key-feature is surely, that both rotation-matrices have a determinant of $1$ ( because $$\small \cos^2(\varphi) + \sin^2(\varphi) = 1 \tag{trigonometric}  $$ and $$\small \cosh^2(\varphi) - \sinh^2(\varphi) = 1 \tag{hyperbolic} $$ ) .        
Now I toyed a bit to extend this to the case of 3-multisection series of the exponential; an example for what I mean is my older question; let's call that three functions just $f(x),g(x),h(x)$ such that $f(x)+g(x)+h(x) = \exp(x)$ and the analogon to the square-formulae which equal $1$ is:
$$ f(x)^3  + g(x)^3 + h(x)^3 - 3f(x)g(x)h(x) = 1 \tag{3-multisection} $$
At least, such a "rotation"-matrix must have size of $3 \times 3$ but possibly even more - if it is constructable at all.
Qu1: Is such a generalization to higher multisections (here order 3) possible?
Qu2: and if, how could such a "rotation"- matrix be contructed?

[Update]: I've just found, that
 $$ \tag{3-multisection} \qquad T_{3m}(\varphi) = \begin{bmatrix} f(\varphi) & h(\varphi) & g(\varphi)\\  g(\varphi)&f(\varphi)&h(\varphi)\\ h(\varphi) & g(\varphi) &f(\varphi)  \end{bmatrix}$$
has determinant $1$ and could be a candidate model. But I didn't find nice properties so far. Perhaps it is even better to not to stick to the determinant 1-condition, but allow determinant $-1$ here; the "rotation"-matrix could then be a simple circulant one, like in the hyperbolic case.      

The log of that matrix looks like
$$ \log(T_{3m}(\varphi))=\small \begin{bmatrix} 
 0 & 0 & \varphi \\ 
 \varphi & 0 & 0 \\ 
 0 & \varphi & 0
 \end{bmatrix}$$
and I think it is a good hint, that this is equivalent to the rotation-matrices in the trigonometric/hyperbolic-cases, where the form of the matrix-log comes out to be much similar.
 A: The analogy I would suggest comes from quadratic forms. Your first two examples are are linear changes of variable that preserve a quadratic form. That is, a matrix $P$ such that $P^T A P = A.$ In the first one, $A= I,$ in the second $A$ is the diagonal matrix with diagonal entries $(1,-1.)$ This is very convenient for homogeneous degree two because of matrices; but we could write 
$$  u = a x + b y, v = c x + d y, u^2 + v^2 = x^2 + y^2 ??  $$ and ask about possible $a,b,c,d.$ Same for $u^2 - v^2 = x^2 - y^2.$
So that suggests $$ u = ax+by+cz, v=dx+ey+fz, w=gx+hy+iz,   $$ what about
$$ u^3 + v^3 + w^3 - 3 u v w = x^3 + y^3 + z^3 - 3 x y z ??    $$
and ask about possible $a,b,c,d,e,f,g,h,i.$ These linear variable changes make a group over the reals. Hmmm. They should, but I need to think about why the determinant of the evident 3 by 3 matrix is nonzero; obvious for the quadratic form case.
EDIT: got it, in the version I am pushing. 
Given  cubic form
$$ f(x,y,z) = x^3 + y^3 + z^3 - 3 x yz. $$
$$ f(x,y,z) =
\det  \left(  \begin{array}{ccc}
  x  &  y & z  \\
  z & x & y \\
   y  &  z  & x
\end{array} 
  \right)  .  $$ 
$f(x,y,z) = \det(xI + y A + z A^2),$ where
$$ A =  \left(  \begin{array}{ccc}
  0  &  1 & 0  \\
  0 & 0 & 1 \\
   1  &  0  & 0
\end{array} 
  \right)    $$ Then $A^3 =  I$ and $A^4 =  A .$ 
Now, since both the characteristic and minimal polynomials for $A$ are $x^3 - 1,$
it follows that all matrices that commute with $A$ are of the form $rI + s A + t A^2.$
A: Isn't this just an element of the Lorenz group from Special Relativity? For Example: $$Q(\beta) = \left[ \begin{matrix} \cosh(\beta) & 0 & 0 & \sinh(\beta) \\
                                    0            & 1 & 0 & 0 \\
                                    0            & 0 & 1 & 0 \\
                                    \sinh(\beta) & 0 & 0 & \cosh(\beta) \end{matrix} \right]$$
