II. (40 points) Find all natural numbers $n$ such that $n \cdot 11^{n} + 1$ is a fifth power number.
Let $n \cdot 11^{n}+1=a^{5}$.
Then when $n=0$, $a=1$.
For $a>1$, we have
$$
n \cdot 11^{n}=(a-1)\left(a^{4}+a^{3}+a^{2}+a+1\right) \text {. }
$$

If $11 \mid (a-1)$, then
$$
\begin{array}{l}
a^{4}+a^{3}+a^{2}+a+1 \equiv 5(\bmod 11) \\
\Rightarrow 11^{n} \mid (a-1), \text { and }\left(a^{4}+a^{3}+a^{2}+a+1\right) \mid n .
\end{array}
$$

Thus, $n \geqslant a^{4}+a^{3}+a^{2}+a+1 > a-1 \geqslant 11^{n} > n$, which is a contradiction.
Therefore, $11 \nmid (a-1)$.
Then $11^{n} \mid \left(a^{4}+a^{3}+a^{2}+a+1\right)$, and
$$
\begin{array}{l}
(a-1) \mid n \Rightarrow n \geqslant a-1 \Rightarrow n+1 \geqslant a \\
\Rightarrow 5(n+1)^{4} \geqslant 5 a^{4} > a^{4}+a^{3}+a^{2}+a+1 \\
\quad \geqslant 11^{n} = (1+10)^{n} .
\end{array}
$$

If $n \geqslant 4$, then
$$
\begin{array}{l}
(1+10)^{n} > \mathrm{C}_{n}^{4} \times 10^{4} \\
\Rightarrow \frac{(1+10)^{n}}{5(n+1)^{4}} > \frac{\mathrm{C}_{n}^{4} \times 10^{4}}{5(n+1)^{4}} \\
\quad = \frac{n}{n+1} \cdot \frac{n-1}{n+1} \cdot \frac{n-2}{n+1} \cdot \frac{n-3}{n+1} \cdot \frac{10^{4}}{120} \\
\quad \geqslant \frac{4}{5} \times \frac{3}{5} \times \frac{2}{5} \times \frac{1}{5} \times \frac{10^{4}}{120} > 1 \\
\Rightarrow (1+10)^{n} > 5(n+1)^{4},
\end{array}
$$

which contradicts equation (2).
Thus, when $n \geqslant 4$, equation (1) has no solution.
Checking $n=1,2,3$, we find that $n=2$ satisfies the requirement.
In conclusion, $n=0$ or 2.