jmc

Let $r$ be an integer root of the first polynomial $p(x)=a_{10}{x}^{10}+a_9{x}^9+a_8{x}^8+\cdots +a_2{x}^2+a_{1}x+a_0=0,$ so $$a_{10}{r}^{10}+a_9{r}^9+\cdots +a_{1}r+a_0=0.$$Since $a_0$ is not equal to 0, $r$ cannot be equal to 0. Hence, we can divide both sides by $r^{10},$ to get $$a_{10}+a_9\cdot \frac{1}{r}+\cdots +a_1\cdot \frac{1}{r^9}+a_0\cdot \frac{1}{r^{10}}=0.$$Thus, $\frac{1}{r}$ is a root of the second polynomial $q(x)=a_0{x}^{10}+a_1{x}^9+a_2{x}^8+\cdots +a_8{x}^2+a_{9}x+a_{10}=0.$ This means that $\frac{1}{r}$ must also be an integer.

The only integers $r$ for which $\frac{1}{r}$ is also an integer are $r=1$ and $r=-1.$ Furthermore, $r=\frac{1}{r}$ for these values, so if the only roots of $p(x)$ are 1 and $-1,$ then the multiset of roots of $q(x)$ are automatically the same as the multiset of roots of $p(x).$ Therefore, the possible multisets are the ones that contain $k$ values of 1 and $10-k$ values of $-1,$ for $0\le k\le 10.$ There are 11 possible values of $k,$ so there are $\boxed{11}$ possible multisets.