Temporarily ignorant?

Here is the table which shows the number of divisors and sum of divisors of numbers from $5$ to $15$ .

$\tau(n)$ represents number of divisors of $n$ and $\sigma(n)$ represents sum of divisors of $n$ . $\begin{array}{c|c|c} n & \tau(n) & \sigma(n) \\ \hline 5 & 2 & 6 \\ 6 & 4 & 12 \\ 7 & 2 & 8 \\ 8 & 4 & 15 \\ 9 & 3 & 13 \\ 10 & 4 & 18 \\ 11 & 2 & 12 \\ 12 & 6 & 28 \\ 13 & 2 & 14 \\ 14 & 4 & 24 \\ 15 & 4 & 24 \end{array}$

1 .Since Nate doesn't knows the number initially, that means $\tau(n$ ) is not unique otherwise he would have known the number.This cancels the possibility of the numbers $9$ and $12$ .

2 .Now Sven also doesn't knows the number. This means that $\sigma(n)$ is also not unique which cancels the possibilities of numbers being $5,7,8,10,12,13$ .

Only possibilities left for $n$ are $6,11,14,15$ .

3 .Now Nate knows the number. This means that he must have got the number $\tau(11)=2$ which is unique rest all are same i.e $4$ .

4 .Now Sven also comes to know that if he (Nate) has known the number, then the number $\tau(n)$ must be unique. Therefore, he also comes to know about the number.

Therefore, only possibility is the number $n=\boxed{11}$