9. (16 points) Given
$$
\begin{array}{l}
f(x)=\log _{a} x(a>0, a \neq 1), \\
g(x)=x^{2}-6 x+9 .
\end{array}
$$

If there exists a unique real number $x$, such that
$$
f(g(x))=g(f(x))=0 \text {, }
$$

find the value of the real number $a$.
9. From $f\left(g\left(x_{0}\right)\right)=0$, we get $g\left(x_{0}\right)=1$. Then $x_{0}^{2}-6 x_{0}+8=0 \Rightarrow x_{0}=2$ or 4. From $g\left(f\left(x_{0}\right)\right)=0$, we get $f\left(x_{0}\right)=3$. Then $\log _{a} 2=3$ or $\log _{a} 4=3 \Rightarrow a=\sqrt[3]{2}$ or $\sqrt[3]{4}$.