Perhaps instead of finite sets one should work with effective divisors, i.e., formal linear combibinations of the form $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bp}{\boldsymbol{p}}$
$$ D= \sum_{\bp\in\bR^N} m_D(\bp) \bp, $$
where $m_D:\bR^N\to\bZ_{\geq 0}$ is a function with finite support. Denote by $\newcommand{\Div}{\mathrm{Div}}$ $\Div_{\geq 0}(\bR^N)$ the space of effective divisors in $\bR^N$. $\Div_{\ge 0}$ has an obvious semigroup structure. Think of the center of mass as a map $\newcommand{\eC}{\mathscr{C}}$
$$ \eC: \Div_{\geq 0}(\bR^N)\to \Div_{\geq 0}(\bR^N), $$
such that for any divisor $D$ the support of the center of math $\eC(D)$ consists of a single point. One important property of this map is
$$ \eC(D_1+D_1) =\eC\Bigl(\; \eC(D_1)+\eC(D_2)\; \Bigr).$$
Another property is
$$\eC(\delta_{\bp})=\delta_{\bp}, $$
where $\delta_{\bp}$ denotes the Dirac divisor of mass $1$ supported at $\bp$.
I believe that these two properties uniquely determine $\eC$. (Maybe some continuity is needed as well.) To extend it to more general mass distributions one needs some continuity assumptions on $\eC$.