The geometric mean of two positive definite (Hermitian) matrices of same size is defined by $$A\natural B := A^{1/2}(A^{-1/2}BA^{-1/2})^{1/2}A^{1/2},$$equivalently, $$A\natural B =(BA^{-1})^{1/2}A=A(A^{-1}B)^{1/2}=\cdots.$$ It is also well-known that $A\natural B$ is the unique positive definite solution $X$ to the Riccati equation $XA^{-1}X=B$, equivalently $X^{-1}AX^{-1}B=I$.
This does not readily generalize to more than two positive definite matrices. But at least for symmetric ones (?? elsewhere this is mentioned without the restriction to symmetric matrices, and that restriction does not seem to be relevant), M. Moakher's article A differential geometric approach to the geometric mean of symmetric positive-definite matrices defines the geometric mean of p.d. matrices $A_1,...,A_m$ as the unique positive definite solution $X$ to the nonlinear matrix equation $$\sum_{j=1}^m\log(X^{−1}A_j)= 0.$$
Of course, we sadly cannot obtain a product of matrices $X^{-1}A_j$ by just applying $\exp$ to that sum if the $A_j$ do not commute and $m\ge3$, but it might be interesting to look, if $X$ denotes their mean, at the volume of the convex hull of all those products $$\Bigl\{ \prod_{j=1}^m(X^{−1}A_{\pi(j)})\Bigl|\pi\in S_m\Bigr\}, $$ which should be relatively "small".
Is it possible to caracterize $m$-tuples of matrices $A_j$ (maybe at least for $m=3$) where this volume is $0$, i.e. the $X^{-1}A_j$ commute?
And in a very different direction:
Can anything be said in the special case that the matrices can be ordered $A_1\le\cdots\le A_m$?
Note that this article has more about the geometric mean of three matrices, especially some geometric background.
