Suppose $X$ is a salient convex cone. That is, if $x,y$$\in$$X$ and $\alpha,\beta$$\geq$$0$ are scalars, then $\alpha$$x$+$\beta$$y$$\in$$X$ and if $0\neq x$$\in$$X$, then $-x$ is not.
Then for any $\bar{x}$$\in$$X$, $$\bigcup_{\sum_{i=1}^{n}{x_i=\bar{x}},x_i\in X}\bigcap_{i=1}^{n}{X+\bar{x}-x_i}=X+(n-1)\bar{x}/n$$ where I use the notation $X+y=${${x+y:x\in X}$}.
Is this claim true? It holds when $X=\mathbb{R}_+$, which I show below.
For $X=\mathbb{R}_+$, take an arbitrary $\bar{x} \in X$ and $x_i \in X$ such that $x_1+...+x_n=\bar{x}$. Then $$\bigcap_{i=1}^{n}{X+\bar{x}-x_i}=[\bar{x}-x_k,\infty)$$ for some $k$ such that $x_k\leq \bar{x}/n$. This implies that $$\bigcap_{i=1}^{n}{X+\bar{x}-x_i} \subseteq X+(n-1)\bar{x}/n$$ and therefore the union is also contained in $X+(n-1)\bar{x}/n$. The reverse inclusion also holds because $\bar{x}/n + ... + \bar{x}/n = \bar{x}$.
Lemma 1: Suppose $X_1,…,X_N$ are convex cones. Then $$\bigcap_{i=1}^{N}{X_i}\subset \sum_{i=1}^{N}{\frac{1}{N}X_i}$$
Proof of Lemma 1: Take $x\in\cap_{i=1}^{N}X_i$. Then $x\in X_i$. Since $X_i$ is a convex cone, $\frac{1}{N}x\in X_i$ for each $i$. Therefore, $$x=\sum_{i=1}^{N}\frac{1}{N}x\in \sum_{i=1}^{N}\frac{1}{N}X_i.\blacksquare $$
Lemma 2: Let $X$ be a convex cone. For any $\bar{x}\in X$,$$\bigcup_{x_1+…+x_N=\bar{x}}_{x_i\in X}\bigcap_{i=1}^{N}X+x_i=X+\frac{1}{N}\bar{x}.$$
Proof of Lemma 2: To prove $(\subset)$, take $x_1,…,x_N$ such that $\sum_{i=1}^{N}{x_i=\bar{x}}$. From Lemma 1, we know that $$\bigcap_{i=1}^{N}{X+x_i}\subset \sum_{i=1}^{N}\frac{1}{N}{(X+x_i)}.$$
Since $X$ is a convex cone, $\frac{1}{N}X=X$. We therefore have that $$\sum_{i=1}^{N}{\frac{1}{N}{(X+x_i)}=X+\sum_{i=1}^{N}{\frac{1}{N}}x_i=X+\frac{1}{N}\bar{x}},$$
since $x_i$ is just a scalar. Since $x_1,…,x_N$ was arbitrary, this holds for all $x_1,…,x_N$ with $x_i\in X$ and $\sum_{i=1}^{N}{x_i}=\bar{x}$. This establishes $(\subset).$
To prove $(\supset)$, suppose $x\in X+\frac{1}{N}\bar{x}$. Let $x_i=\frac{1}{N}x$. Then $\sum_{i=1}^{N}{\frac{1}{N}x_i}=x$, and $\frac{1}{N}x\in X+\frac{1}{N}x$, establishing the claim.$\blacksquare$
Lemma 2 applies to all convex cones, including salient convex cones (which is the non-trivial case). The claim in the question is then a straightforward corollary (involving minor relabeling).