nt.number theory – Does ${tau(1)tau(2)+cdots+tau(n-1)tau(n)+tau(n)tau(1): tauin S_n}$ contain a unique multiple of $n^2$ for each $nge6$?

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

Question.
Does the set $$T_n:={tau(1)tau(2)+cdots+tau(n-1)tau(n)+tau(n)tau(1): tauin S_n}$$ contain a unique multiple of $n^2$ for each $nge6$?

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

For $n=7$, we have
$$1times3+3times4+4times5+5times6+6times2+2times7+7times1=2times7^2.$$
For $n=8$, we have
$$1times5+5times3+3times6+6times4+4times7+7times2+2times8+8times1=2times8^2.$$
For $n=9$, we have
$$1times2+2times3+3times5+5times4+4times6+6times8+8times7+7times9+9times1=3times9^2.$$
For $n=10$, we have
begin{gather*}1times2+2times3+3times6+6times8+8times4+4times9+9times7+7times5+5times10+10times1
\=3times10^2.end{gather*}

nt.number theory – Is there a permutation $tauin S_n$ with $tau(1)^{tau(2)}+cdots+tau(n-1)^{tau(n)}+tau(n)^{tau(1)}$ a square?

Let $n>1$ be an integer, and let $S_n$ be the symmetric group of all the permutatins of ${1,ldots,n}$.
I’m curious whether there is a permutation $tauin S_n$ such that
$$tau(1)^{tau(2)}+cdots+tau(n-1)^{tau(n)}+tau(n)^{tau(1)}$$
is a square. (Without loss of generality we may assume that $tau(n)=n$.) For $n=2,3$ there is no such a permutation $tau$. But my computations for $n=4,5,ldots,11$
lead me to formulate the following conjecture.

Conjecture. For any integer $n>3$, there is a permutation $tauin S_n$ such that
$$tau(1)^{tau(2)}+cdots+tau(n-1)^{tau(n)}+tau(n)^{tau(1)}$$
is a square.

For example,
$$2^1 + 1^3 + 3^4 + 4^2 = 10^2, 1^5 + 5^2 + 2^4 + 4^3 + 3^6 + 6^1 = 29^2,$$
and
$$1^3 + 3^2 + 2^{10} + 10^5 + 5^7 + 7^8 + 8^6 + 6^9 + 9^4 + 4^{11} + 11^1 = 4526^2.$$
For more examples and related data, one may consult http://oeis.org/A342965.

QUESTION. Is the above conjecture true?

Your are welcome to check the conjecture further.