jmc

By the Cauchy-Schwarz inequality, $$\begin{array}{rl}& [(1^2+1^2+1^2+\cdots +1^2+1^2)][(1-x_1{)}^2+(x_1-x_2{)}^2+(x_2-x_3{)}^2+\cdots +(x_9-x_{10}{)}^2+x_{10}^2]\\ & \ge [(1-x_1)+(x_1-x_2)+(x_2-x_3)+\cdots +(x_9-x_{10})+x_{10}{]}^2=1.\end{array}$$From the given condition, we have equality, so by the equality condition for Cauchy-Schwarz, $$\frac{1-x_1}{1}=\frac{x_1-x_2}{1}=\frac{x_2-x_3}{1}=\cdots =\frac{x_9-x_{10}}{1}=\frac{x_{10}}{1}.$$Let $$d=1-x_1=x_1-x_2=x_2-x_3=\cdots =x_9-x_{10}=x_{10}.$$Then $$(1-x_1)+(x_1-x_2)+\cdots +(x_9-x_{10})+x_{10}=11d,$$so $11d=1.$ Then $d=\frac{1}{11},$ so $$(x_1,x_2,x_3,\dots ,x_{10})=\left(\frac{10}{11},\frac{9}{11},\frac{8}{11},\dots ,\frac{1}{11}\right).$$In particular, there is only $\boxed{1}$ solution.