I am not 100% about this claim, so I would like a proof or a counterexample.
First, let $\mathbb{R}^n$ have standard basis $e_1,\dots, e_n$ and $x\in\mathbb{R}^n$ have coordinates $(x_1,\dots, x_n)$. Suppose $Q$ is a multiple of $e_1$ and let $\mu$ be the measure on $\mathbb{R}^n$ with density $(1+|x-Q|^2)^{-(n+1)/2}$. Let $S$ be a simplex with vertices $s_0\in\mathbb{R}^n$ and $s_1,\dots, s_n\in Q^\perp$.
Claim: If $T$ is an linear transformation fixing $Q^\perp$, then
$$0=\int_S (x-Q)d\mu=\int_{TS} (x-Q)d\mu$$
implies $TS=S$.
Progress: Since $T$ is linear and fixes $e_2,\dots, e_d$, there must be some $\tau\in\mathbb{R}^n$ so that $$T(x)=x+x_1\tau$$ The premise becomes
$$\int_S (x-Q)(1+|x-Q|^2)^{-(n+1)/2}dx=\int_{TS} (x-Q)(1+|x-Q|^2)^{-(n+1)/2}dx$$
Now, enact a change of variables to get the regions to match,
$$\int_S (x-Q)(1+|x-Q|^2)^{-(n+1)/2}dx=\int_{S} (x+x_1\tau-Q)(1+|x+x_1\tau-Q|^2)^{-(n+1)/2}dx$$
I expect since $T$ moves all the points in $S$ in the direction of $\tau$ that the right hand integral in the claim will also move in the direction of $\tau$ (specifically a positive inner product with $\tau$).
We might suppose $\tau\neq 0$ and take inner product with $\tau$ on both sides to maybe get inequality?
Thanks for looking!