3. Given the function $f(x)=x^{2}-2 x+a$. If $\{x \mid f(x)=x\}=\{x \mid f(f(x))=x\}$, then the range of the real number $a$ is $\qquad$
3. $\left[\frac{5}{4},+\infty\right)$.

From $f(f(x))=x$
$$
\begin{array}{l}
\Rightarrow f(f(x))-f(x)=x-f(x) \\
\Rightarrow(f(x)-x)(f(x)+x-1)=0 .
\end{array}
$$

According to the problem, we know that the equation $f(x)+x-1=0$ has no real roots or has the same roots as the equation $f(x)-x=0$.
Then $\Delta=5-4 a<0$ or $f\left(\frac{1}{2}\right)=\frac{1}{2}$.
Solving this, we get $a \geqslant \frac{5}{4}$.
Therefore, the range of real number $a$ is $\left[\frac{5}{4},+\infty\right)$.