Is there a reasonable characterization of all the "power-law" diffeomorphisms of finite order $f:\mathbb R^{>0} \times \mathbb R^{>0} \to \mathbb R^{>0} \times \mathbb R^{>0}$, of the form $f(x,y)=(x^ky^r,x^{1-k}y^{1-r})$?
i.e. I am looking for (homogeneous powers) $f=(f_1(x,y),f_2(x,y))$ of finite order, such that
$$ f_1(x,y)\cdot f_2(x,y)=xy \, \, \text{ for every } \, \, x,y >0.$$
If we use the notation $(k,r)$ to encode $f$, then $f^2=(k^2+r-kr,kr+r-r^2)$. A short analysis shows that $f$ is of order $2$ exactly when $k=r-1$.
Is there any tractable way to push this reasoning forward, to higher orders?
Believe it or not, this is a linear algebra problem: take logs of everything in sight, and your scenario reduces to analyzing the map $\Bbb R^2\to \Bbb R^2$ given by the matrix $$\begin{pmatrix} k & r \\ 1-k & 1-r \end{pmatrix}$$ and asking when it is of finite order.
The eigenvalues of this matrix are $1$ and $k-r$, so the only possible way that this is a matrix of finite order is if $k-r=\pm 1$. If $k-r=1$, then this matrix is either the identity ($r=0$) or similar to a size-two Jordan block ($r\neq 0$). If $k-r=-1$, then this is the case you've already identified, and the matrix is of order 2.
In particular, every map with $k=r-1$ is of order two, and there are no other non-identity "power-law maps" with finite order. So this is the complete classification.