11. Let $f(x)=x^{2}+a x+b(a, b \in \mathbf{R}), A=\{x \mid f(x)=x, x \in \mathbf{R}\}$, $B=\{x \mid f(f(x))=x, x \in \mathbf{R}\}$. If $A=\{-1,3\}$, then $B=$ $\qquad$
11. $\{-1, \sqrt{3},-\sqrt{3}, 3\}$.

From $A=\{-1,3\}$, we get $x^{2}+a x+b-x=(x+1)(x-3), a=-1, b=-3$, then the equation $f(f(x))=x$ becomes $\left(x^{2}-x-3\right)^{2}-\left(x^{2}-x-3\right)-3=x$, solving this yields $B=\{-1, \sqrt{3},-\sqrt{3}, 3\}$.