Minimum sum of the squares

Building off of gammatester's counterexample, it looks like a correct conjecture might be to run up the odds and down the evens, e.g.,

$$1,3,5,7,9,10,8,6,4,2$$

Note that no matter how you arrange the numbers, the average difference between (circularly) consecutive numbers is always $0$, so in a sense what you're trying to do is find an arrangement that minimizes the variance. I.e., if you fix a circular arrangement and let $X$ be the random variable for the difference between a randomly chosen pair of consecutive numbers, then $V(X)=E(X^2)-E(X)^2=E(X^2)-0$.

Your hypothesis is wrong, look e.g. at $n=4$. Your formula will give the sum $3\times 4 = 12$. But if you take $x_1\dots x_4 = 1,3,4,2\;$ the sum is $2^2 + 1^2 + 2^2 + 1^2 = 10$