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 $\newcommand{\eC}{\mathscr{C}}$

$$ \eC: \Div_{\geq 0}(\bR^N)\to \Div_{\geq 0}(\bR^N),  $$

with the following properties. 

**1.** For any divisor $D$,  the support of the center of mass $\eC(D)$ consists of a single point.  

**2.** For any divisors $D_1,D_2\in\Div_{\geq 0}(\bR^N)$  we have 

$$ \eC(D_1+D_1) =\eC\Bigl(\; \eC(D_1)+\eC(D_2)\; \Bigr).$$

**3.** For any point $\bp\in \bR^N$ and any $m\in\bZ_{\geq 0}$

$$\eC(m \delta_{\bp})=m\delta_{\bp}, $$

where $\delta_{\bp}$ denotes the Dirac divisor of mass $1$ supported at  $\bp$.

**4.** $\eC$ is continuous with respect to the obvious topology on $\Div_{\geq 0}(\bR^N)$, where supports of divisors converge in the Hausdorff metric and in the limit the total mass is conserved. 

**Claim.** I believe that these properties uniquely determine  $\eC$.  

**Weaker Claim.**  The map $\eC$ is uniquely determined if  besides  **1**,...**4** we assume the following additional condition.

**5.** If the support of $D$ is contained in an affine subspace $V\subset \bR^N$, then  the support of $\eC(D)$ is contained in the same affine subspace.


**Edit 1.** Strengthened  Condition $3$.  The function

$$  \Div_{\geq 0}(\bR^N)\ni D=\sum_{\bp} m(\bp)\delta_{\bp}\mapsto \eC_0(D):=m(D)\delta_{C(D)} \in  \Div_{\geq 0}(\bR^N)\tag{$\ast$}, $$

where 

$$ m(D) :=\left(\sum_{\bp} m(\bp)\right) ,\;\; c(D)=  \frac{1}{m(D)}\sum_{\bp} m(\bp) \bp, $$



satisfies  all the above  $5$ conditions.   The Weak Claim states  that  this is the only function satisfying these conditions.

**Edit 2.**    I took  into account    Deanne's comments  and I was able to prove  the following version of the  above claim.

**PROPOSITION.**  There exists exactly one map $\eC:\Div_{\geq 0}(\bR^N)\to \Div_{\geq 0}(\bR^N)$ satisfying the following conditions.

**(Localization.)**  The support of $\eC(D)$ consists of a single point.

**(Conservation of mass.)** 

$$ m\bigl( D\bigr)=m\bigl(\;\eC(D)\;\bigr),\;\;\forall  D\in \Div_{\geq 0}$$ 

**(Normalization.)**$\newcommand{\bq}{\boldsymbol{q}}$   $$\eC(m\delta_{\bp}) =m\delta_{\bp},\;\; \eC(\delta_{\bp}+\delta_{\bq})=2\delta_{\frac{1}{2}(\bp+\bq)}. $$

**(Additivity.)** 

$$\eC(D_1+D_2)=\eC\bigl(\;\eC(D_1)+\eC(D_2)\;\bigr),\;\;\forall D_1,D_2\in\Div_{\geq 0}. $$

[You can find a proof **here**.][2]

**Edit 3.**  The [proof][1] of the above proposition   does not use the linear structure of $\bR^N$. It uses only the *metric structure*, and  more precisely the fact that  any two different points in $\bR^N$ determine a *unique* geodesic.   The normalization condition  should be rewritten as

$$\eC(\delta_{\bp}+\delta_{\bq})= 2 \delta_{c(\bp,\bq)}, $$

where $c(\bp,\bq)$ denotes the midpoint of the geodesic arc $[\bp,\bq]$.   In the above proposition we can then replace $\bR^N$ with the hyperbolic space $\newcommand{\bH}{\boldsymbol{H}}$ $\bH^N$  and we can conclude that on a hyperbolic space there exists  at most one notion of center of mass, i.e., a map $\eC:\Div_{\geq 0}(\bH^N)\to   \Div_{\geq 0}(\bH^N)$  satisfying the four conditions in the Proposition.




  [1]: http://liviusmathblog.blogspot.com/2012/12/axiomatic-definition-of-center-of-mass.html
  [2]: http://liviusmathblog.blogspot.com/2012/12/axiomatic-definition-of-center-of-mass.html