10. If there exists a unique positive integer $n$ satisfying:
$$
\log _{2} \log _{16} n=\log _{4} \log _{4} n,
$$

then the sum of the digits of $n$ is ( ).
(A) 4
(B) 7
(C) 8
(D) 11
(E) 13
10. E.

From the problem, we know
$$
\begin{array}{l}
\log _{2} \log _{16} n=\log _{4} \log _{4} n \\
\Rightarrow \log _{2}\left(\frac{1}{4} \log _{2} n\right)=\frac{1}{2} \log _{2}\left(\frac{1}{2} \log _{2} n\right) \\
\Rightarrow-2+\log _{2} \log _{2} n=\frac{1}{2}\left(-1+\log _{2} \log _{2} n\right) \\
\Rightarrow-4+2 \log _{2} \log _{2} n=-1+\log _{2} \log _{2} n \\
\Rightarrow \log _{2} \log _{2} n=3 \\
\Rightarrow \log _{2} n=8 \\
\Rightarrow n=2^{8}=256 .
\end{array}
$$

Therefore, the sum of the digits of $n$ is 13.