Nonlinear operator monotone bijections on the set of real positive semidefinite $d \times d$ matrices

Question

I was wondering if there are any nonlinear operator monotone (with respect to the Loewner order) bijections on the set of real symmetric positive semidefinite matrices $S_+^d$, since I could only come up with linear ones, e.g. $X \mapsto U^T X U$, where $U$ is orthogonal (are those all linear operator monotone bijectins on $S_+^d$?). These linear bijections are convex, but are there also nonlinear operator monotone convex bijections on $S_+^d$ (perhaps in some sense generalising the power functions $x \mapsto x^p$, $p \in [1, \infty)$?)

Unfortunately, extending an operator monotone bijection to matrices does not yield an operator monotone matrix function (e.g. $x \mapsto x^2$), where $f \colon S_+^d \to S_+^d$ being operator monotone means $X \preceq Y$ $\implies$ $f(X) \preceq f(Y)$ and $A \preceq B$ is defined via $B - A \in S_+^d$.

The matrix logarithm and exponential are not bijective. In the comments it was mentioned that the square root is nonlinear and operator monotone, but it is non-convex.

The motivation is me trying to generalise of a certain optimization problem from measures with values in $[0, \infty)$ to tensor-valued measures, that is, measures having as values real symmetric positive definite matrices. In that paper, a operator monotone bijection (in this case $x \mapsto x^2$ is used) and I want to find a suitable alternative.

Answer 1

Thanks to the paper linked in the comments by @mechanodroid, I can answer / comment on some of the questions I raised.

On page 39, it is stated that (adapted to our case)

As the theory of operator monotone functions is highly non-trivial there is no much hope to get a nice description of bijective maps on $S_+^d$ that preserve order in one direction only.

Further, (p. 4) in our case if $d \ge 2$ and $f \colon S_+^d \to S_+^d$ is bijective and $X \le Y$ $\iff$ $f(X) \le f(Y)$ ($f$ is an order isomorphism), then $f(A) = T A T^T$ for some invertible matrix $T$ (the reverse implication also holds).

In particular: an order isomorphism has to be linear (and thus convex).

Note also that the power functions $f_p \colon [0, \infty) \to [0, \infty)$, $x \mapsto x^p$ have the property $$x < y \iff f_p(x) < f_p(y),$$ so it might be instructive that a generalisation of this power functions is a order isomorphism (instead of an operator monotone bijection).