Question
Let's say I have $n$ fair 6-sided dice. What is the probability that the same set of outcomes is obtained when $n$ fair dice are rolled twice?
Example
$n=8$: Suppose $8$ dice are rolled. If the first roll is $[1, 3, 6, 2, 3, 4, 3, 1]$, there are two 1's, one 2, three 3's, one 4, and one 6.
Rolling the same dice again, what is the probability that the next roll will also have two 1's, one 2, three 3's, one 4, and one 6?
The above example uses 8 dice, but I'm curious about the probability for any positive $n$.
Edit
My question is different than Two dice throw probability since I'm looking for the probability of $n$ dice having the same result in two consecutive throws
You have multinomial distribution with $n$ trials, $k = 6$ outcomes and all $p_i = \frac{1}{6}$. So probability of getting two equal rolls is $\sum\limits_{x_1 + \ldots + x_6 = n} \left(\frac{n!}{x_1! \ldots x_6!}\cdot \frac{1}{6^n}\right)^2$. It's unlikely there is good closed form for it. For asymptotic see, for example, this answer on mathoverflow.
For example, if we have $n = 2$, we need to sum over all variants to partitioning $2$ into $6$ non-negative terms. There are $15$ of them with $2$ ones and $4$ zeroes: $1+1+0+0+0+0$, $1+0+1+0+0+0$, ..., $0+0+0+0+1+1$ and $6$ with $1$ two and $5$ zeroes: $2+0+0+0+0+0$, $0+2+0+0+0+0$, ...
First variant will give in our sum $15$ terms equal to $\left(\frac{2!}{1!1!0!0!0!0!} \cdot\frac{1}{36}\right)^2 = \frac{1}{18^2}$. Second will give $6$ terms equal to $\left(\frac{2!}{2!0!0!0!0!0!} \cdot \frac{1}{36^2}\right)^2 = \frac{1}{36^2}$. So the answer is $\frac{1}{18^2}\cdot 15 + \frac{1}{36^2}\cdot 6 \approx 0.05$.