9.7.

Notice that, $2015=5 \times 13 \times 31$.
Therefore, 2015 has eight positive divisors.
Let $G(n)$ denote the unit digit of $n$, then
$$
\begin{array}{l}
G\left(1^{2015}\right)=G\left(31^{165}\right)=1, \\
G\left(2015^{1}\right)=G\left(5^{403}\right)=G\left(65^{31}\right) \\
=G\left(155^{13}\right)=5 .
\end{array}
$$

From $G\left(3^{4}\right)=1$, we know
$$
\begin{array}{l}
G\left(13^{155}\right)=G\left(3^{155}\right)=G\left(3^{4 \times 38+3}\right) \\
=G\left(3^{3}\right)=7, \\
G\left(403^{5}\right)=G\left(3^{5}\right)=3 .
\end{array}
$$

Thus, the maximum value of the unit digit of $d^{\frac{2015}{d}}$ is 7.
9. Let $d$ be a positive divisor of 2015. Then the maximum value of the unit digit of $d^{\frac{2005}{d}}$ is $\qquad$ .