Probability of obtaining equal number of each outcome of a fair die at the nth trial

Question

Suppose a fair die is tossed repeatedly. I am concerned in deriving the probability of the occurrence of obtaining equal number of each possible outcome at the nth trial. Clearly this is only possible for multiples of 6 hence this has a period of 6. I need to obtain a formula for $u_n$ (the probability of this event occurring at the nth trial but not necessarily for the first time) Is it by any chance $(1/6)^{6n}$ ? Any assistance will be much appreciated. Thank you

Answer 1

$$u_{6n}=\frac{(6n)!}{(n!)^6}\frac1{6^{6n}}\sim\frac{\sqrt{3\pi}}{4\pi^3}\frac1{n^2\sqrt{n}}$$

Answer 2

For $n=6m$, the probability is: $$\frac{\prod\limits_{k=0}^{5}\binom{n-km}{m}}{6^n}$$