Round about

We will ignore the condition that $x_1$ is the minimum, and apply it right at the end.

First we will show that all $x_i$ are between $-2$ and $2$ . Clearly, $x_1=2-x_4^4\leq 2,$ and similarly for the other $x_i.$ To show that $x_1$ cannot be less than $-2,$ we will use the following simple lemma.

Lemma. If $a<-2,$ then $2-a^2<a$ .

Proof. Because $a^2+a-2=(a-1)(a+2),$ it is positive for $a<-2,$ which implies the desired inequality.

Now suppose that $x_1<-2.$ Applying the above lemma four times, we get $x_1>x_2>x_3>x_4>x_1,$ a contradiction.

So $-2\leq x_1 \leq 2.$ Therefore we can write $x_1=-2\cos t.$ Therefore, $x_2=2-x_1^2=-2(2\cos ^2 t-1)=-2\cos 2t.$ Similarly, $x_3=-2\cos 4t,$ $x_4=-2\cos 8t,$ $x_1=-2\cos 16t.$ So $\cos t =\cos 16t,$ and this condition is both necessary and sufficient.

This is equivalent to $16t=\pm t + 2\pi k,$ for some integer $k.$ We will distinguish two cases.

Case 1. $16t=t+2\pi k.$ Then $15t=2\pi k,$ $t=\frac{k}{15}\cdot 2\pi.$ Clearly, it is enough to consider $k=0,1,2,...,14$ to get all possible values of $\cos t.$ Moreover, $\cos t =\cos (2\pi -t),$ so it is enough to consider $k=0,1,...,7.$

If $k=0,$ then $2k=4k=8k=0,$ we will write this as $(0,0,0,0).$ This does not satisfy our condition that the four numbers are distinct.

If $k=1,$ then we get $(1,2,4,7)$ (note that $8=15-7$ . Similarly, if $k=2,$ we get $(2,4,7,1).$ We also get $(4,7,1,2)$ and $(7,1,2,4).$ These lead to the same set of four distinct numbers, so we have found one round set.

If $k=3,$ we get $(3,6,3,6),$ so the four numbers are not distinct. If $k=5,$ we get $(5,5,5,5),$ which is also not allowed. Since all values of $k$ have been accounted for, there is exactly one round set coming from this case.

Case 2. $16t=-t+2\pi k.$ Then $17t=2\pi k,$ $t=\frac{k}{17}\cdot 2\pi.$ As in Case 1, we can restict our attention to $k=0,1,2,...,16$ and then to $k=0,1,...,8.$ A simple analysis, like in Case 1, gives the following solutions:

$(0,0,0,0)$ discarded because the numbers are the same (it is also the same as the $(0,0,0,0)$ from Case 1, it corresponds to $x_1=x_2=x_3=x_4=-2$ ).

$(1,2,4,8),$ $(2,4,8,1),$ $(4,8,1,2),$ $(8,1,2,4).$ These give the same round set.

$(3,6,5,7),$ $(6,5,7,3),$ $(5,7,3,6),$ $(7,3,6,5).$ This is another round set.

So, altogether there are $3$ round sets.