The answer is $21^5$. Let $A_i=\{k|k\equiv i(mod\ 5),1\leq k\leq 100\},i=1,2,3,4$. First ,fix $i$ and we notice that as one transformation is done,the number of $1$s in the $x_k(k\in A_i)$ doesn't change.Then by choosing different number of $1$s in each $A_i$ ,we can get $21^5$ different numbers satisfying the condition. And on the other hand ,we have it that we can transform $\overline{x_0x_1x_2x_3x_4x_5x_6x_7x_8x_9x_{10}}$ into $\overline{x_5x_6x_7x_8x_9x_0x_1x_2x_3x_4x_{10}}$ and then into $\overline{x_5x_1x_2x_3x_4x_{10}x_6x_7x_8x_9x_0}$ ,which means that we can change the order of $3$ consecutive digits in a fixed $A_i$ without affecting other $A_i$s.With this we can easily find out that we can change the order of the whole $A_i$ without affecting other $A_i$s.And this implies that $21^5$ is the best number for the problem.