Maximal number of $S_n$-conjugates living in a hyperplane

Let $$v=(a_1,\dots,a_n)\in\mathbb{R}^n$$ where the $$a_i$$ are distinct and positive. For $$\sigma\in S_n$$, let $$\sigma(v)=(a_{\sigma(1)},\dots,a_{\sigma(n)})$$. For any hyperplane $$H$$ through the origin, let $$m_H$$ be the number of $$\sigma\in S_n$$ such that $$\sigma(v)\in H$$. What is $$\max_H m_H$$?

I suspect that the answer is $$(n-1)!$$. Notice that if $$\sigma\in S_{n-1}$$, then the $$(n-1)!$$ vectors $$\sigma(v)$$ are all contained in the hyperplane $$(\sum_{i=1}^{n-1}a_i)x_n=a_n\sum_{i=1}^{n-1}x_i$$.

• Do you mean the maximum over all possible choices of $v$, or the maximum as a function of $v$? (Also, are your hyperplanes always through the origin?) – LSpice Nov 20 '18 at 0:33
• Good point. I edited the question. – user131566 Nov 20 '18 at 1:01
• If $n$ is even and $v=(1,2,\dots,n)$, then the condition $\frac{n-2}2(x_1+x_2)=(x_3+\dots+x_n)$, equivalent with $x_1+x_2=n+1$, is satisfied on $(n-2)!n$ points. – Ilya Bogdanov Nov 20 '18 at 10:35
• Your edit still doesn't seem directly to address my question: are you looking for $\max_H m_H(v)$ as a function of $v$, or for $\max_v \max_H m_H(v)$? – LSpice Nov 20 '18 at 15:28
• Understanding $\max_H m_H(v)$ would be ideal, but I would be happy even knowing the answer to $\max_v\max_H m_H(v)$. – user131566 Nov 20 '18 at 15:32