$H$ is a matrix that defines hyperbolic rotation in the standard form. This essentially means we are sliding all points along the hyperbola by some angle.
$$H=\exp\begin{pmatrix} 0 & s \\ s & 0\end{pmatrix} = \begin{pmatrix} \cosh s & \sinh s \\ \sinh s & \cosh s\end{pmatrix}.$$
Similarly for a rectangular hyperbola $B$ gives the hyperbolic rotation and we are essentially sliding all points along the hyperbola $f(x)=\frac{1}{x}.$ The transformation matrix that achieves this is
$$A=\begin{pmatrix} e^s & 0 \\ 0 & e^{-s}\end{pmatrix}.$$
Given a function like $g(x)=\exp(\frac{1}{\log(x)})$ how can I find the transformation matrix that slides all points along $g?$
I could be wrong but I think it is simply $$R=\exp(A).$$ for $x,g>1.$