Does $\{\tau(1)\tau(2)+\cdots+\tau(n-1)\tau(n)+\tau(n)\tau(1):\ \tau\in S_n\}$ contain a unique multiple of $n^2$ for each $n\ge6$?

Question

Motivated by Question 397575, here I pose a related question.

Question. Does the set $$T_n:=\{\tau(1)\tau(2)+\cdots+\tau(n-1)\tau(n)+\tau(n)\tau(1):\ \tau\in S_n\}$$ contain a unique multiple of $n^2$ for each $n\ge6$?

I conjecture that the answer is positive. I have verified this for $n=6,\ldots,10$. For $n=6$, we have \begin{align*}&2\times4+4\times1+1\times3+3\times5+5\times6+6\times2 \\=&3\times5+5\times1+1\times2+2\times4+4\times6+6\times3=2\times6^2. \end{align*} For $n=7$, we have $$1\times3+3\times4+4\times5+5\times6+6\times2+2\times7+7\times1=2\times7^2.$$ For $n=8$, we have $$1\times5+5\times3+3\times6+6\times4+4\times7+7\times2+2\times8+8\times1=2\times8^2.$$ For $n=9$, we have $$1\times2+2\times3+3\times5+5\times4+4\times6+6\times8+8\times7+7\times9+9\times1=3\times9^2.$$ For $n=10$, we have \begin{gather*}1\times2+2\times3+3\times6+6\times8+8\times4+4\times9+9\times7+7\times5+5\times10+10\times1 \\=3\times10^2.\end{gather*}

Answer 1

No, for $n = 11$ this fails:

363 = 3 * 11^2 with [7, 2, 8, 5, 3, 4, 6, 9, 10, 1, 11]

484 = 2^2 * 11^2 with [10, 9, 6, 3, 1, 2, 4, 5, 7, 8, 11]

Running the code I wrote to check this a little more, there is more than one multiple of $n^2$ in the set you describe for all $11\leq n \leq 50$, see here for two permutations leading to different multiples of $n^2$ for each such $n$.