$$
\begin{array}{l}
\text { 11. Let } f(x)=x^{2}+a x, \\
\{x \mid f(x)=0, x \in \mathbf{R}\} \\
=\{x \mid f(f(x))=0, x \in \mathbf{R}\} \neq \varnothing .
\end{array}
$$

Then the range of all real numbers $a$ that satisfy the condition is
11. $\{a \mid 0 \leqslant a<4\}$.

Since the roots of $f(x)=0$ are $x=0$ or $x=-a$, we can transform $f(f(x))=0$ into
$$
f(x)=0 \text { or } f(x)+a=0 \text {. }
$$

From the problem, $f(x)+a=0$ has no solution or the roots of $f(x)+a=0$ are $x=0$ or $x=-a$.
And $f(x)+a=0$, which is $x^{2}+a x+a=0$.
Thus, $\Delta=a^{2}-4 a<0$ or $a=0$.
Therefore, $0 \leqslant a<4$.