Let $O$ be the origin, $\overline{OA_i}=a_i$, $|a_i|=1$, $\overline{OO_i}=p_i$, $\overline{OG}=(\sum a_i)/(n+1)=:a$. We want to prove that $\sum p_i=0$. We know that $(p_i-a)^2=(p_i-a_j)^2$ for $j\ne i$. That is, $(p_i,a_j)=(p_i,a)+a_j^2/2-a^2/2$ for $j\ne i$. Sum up by all $j\ne i$, we get $(p_i,(n+1)a-a_i)=n(p_i,a)+n/2-na^2/2$. Thus $(p_i,a_i)=(p_i,a)-n/2+na^2/2$. Therefore $(\sum_i p_i,a_j)=(p_j,a_j)+\sum_{k\ne i}(p_k,a_j)=(\sum_i p_i,a)$, $\sum_i p_i$ is orthogonal to $a-a_i$, but these vectors generate the whole space and we conclude that $\sum_i p_i=0$.

Let $O$ be the origin, $\vec{OA_i}=a_i$, $\vec{OO_i}=p_i$, $\vec{OG}=a$.

Then $\|a_i\|=1$ and $\sum a_i/(n+1)=a$. We want to prove that $\sum p_i=0$. 

We know that for $i \neq j$: \begin{align} \|p_i-a\|^2 &= \|p_i-a_j\|^2\\ (p_i,a_j) &= (p_i,a)+\frac{\|a_j\|^2-\|a\|^2}{2}. \end{align} Summing up over all $j\ne i$: \begin{align} (p_i,(n+1)a-a_i) &= n(p_i,a)+\frac{n-n\|a\|^2}{2} \\ (p_i,a_i) &= (p_i,a)+\frac{-n+n\|a\|^2}{2} \\ \sum_i(p_i,a_j) &= (p_j,a_j)+\sum_{k\ne i}(p_k,a_j)=\sum_i (p_i,a). \end{align} Now $\sum_i p_i$ is orthogonal to $a-a_i$, but these vectors generate the whole space and we conclude that $\sum_i p_i=0$.