Say we have a finite random vector of exchangeable variables $X =(X_{1},X_{2},\dots,X_{m}) $, i.e., $(X_{1},X_{2},\dots,X_{m}) \stackrel{d}{=} (X_{\pi(1)},X_{\pi(2)},\dots,X_{\pi(m)})$ for any finite permutation $\pi$ of the indices $1, 2, \dots,m$.
Now assume linear map $L_{i}:R^m\mapsto R^{m-1}$ such that $Y_{i}=L_{i}X=(X_{1}-X_{i},X_{2}-X_{i},\dots,X_{m}-X_{i} )$, that is $Y_{i} = (X_{l}-X_{i}, l=1,\dots,m, l\neq i )$.
My question is, what can we say about $Y_{i}$ and more precisely about its distribution and support ?
What I can see is that $Y$ is still exchangeable since $Y_{i}\stackrel{d}{=}(X_{\pi(1)}-X_{i},X_{\pi(2)}-X_{i},\dots,X_{\pi(m)}-X_{i} ) $ (where $\pi$ this time maps only to $\{1\dots,m\}\backslash\{i\}$), however for $m=2$ there is more to observe. Set $m=2$ and $i=2$ then $Y_{2}=(X_{1}-X_{2})\stackrel{d}{=}(X_{2}-X_{1})=-Y_{2}$ thus $Y_{2}$ has a symmetric distribution about 0.
Can this phenomena be seen also in the case $m>2$? Some symmetry that must be satisfied by the distribution or by the support?
If we assume that the following expectations exists we conclude that from the permutation symmetry
$$ E \left ( \prod_{i=1}^k f_i(X_i)^{n_i} \right ) = E \left ( \prod_{i=1}^k f_i(X_{\pi(i)})^{n_i} \right ) $$
So especially all moments of $X_i-X_j$ is the same as all moments of $X_j-X_i$ which means that in this case they are symmetric. Likewise your assumption on $Y$ is correct. To answer generally note that all your r.v is linear combinations and use $e^{ax+by} = e^{ax}e^{by}$ and characteristic function,