For a particular solution $(v_1, v_2, \dots, v_{10})$, let $s = v_1^2 + v_2^2 + \dots + v_{10}^2$.
Then
$$v_i = 1 + \frac{6v_i^2}{s} \quad \Rightarrow \quad 6v_i^2 - sv_i + s = 0.$$
Let $a$ and $b$ be the roots of the quadratic $6x^2 - sx + s = 0$, so for each $i$, $v_i = a$ or $v_i = b$. We also have $ab = s/6$ (by Vieta's formula, for example).
If all the $v_i$ are equal, then
$$v_i = 1 + \frac{6}{10} = \frac{8}{5}$$
for all $i$. Otherwise, let $5+k$ of the $v_i$ be $a$, and let $5-k$ of the $v_i$ be $b$, where $0 < k \le 4$. Then by the AM-GM inequality,
$$6ab = s = (5 + k)a^2 + (5 - k)b^2 \ge 2ab\sqrt{25 - k^2}.$$
From the given equations, $v_i \ge 1$ for all $i$, so $a$ and $b$ are positive.
Then $\sqrt{25 - k^2} \le 3 \Rightarrow 25 - k^2 \le 9 \Rightarrow k^2 \ge 16 \Rightarrow k = 4$. Hence,
$6ab = 9a^2 + b^2 \Rightarrow (b - 3a)^2 = 0 \Rightarrow b = 3a$.
Adding all given ten equations, we get
$$v_1 + v_2 + \dots + v_{10} = 16.$$
But $v_1 + v_2 + \dots + v_{10} = 9a + b = 12a$, so $a = 16/12 = 4/3$ and $b = 4$. Therefore, the solutions are $(8/5, 8/5, \dots, 8/5)$ and all ten permutations of $(4/3, 4/3, \dots, 4/3, 4)$.