Incorrect answer. Please look at #3, #4.

youthdoo wrote: Let the smallest number among the $n$ be $x$. Firstly, $n=9$ doesn't work because unless all the digits of $x$ are $1$, $9x$ has more digits than $x$, which is contradictory. However, $11\dots1$ clearly doesn't meet the requirements.

For $n=8$, we have $x=1176470588235294$. How to get this

So the largest $n$ is $n=8$.

Verification by python

Try 2/19

Thanks for noticing. For $n=9$, we have $$x=105263157894736842.$$(This is obtained from the repetend of $\frac2{19}$.) The numbers $x,2x,\dots,9x$ meet the requirements. For $n\ge10$, the greatest number is at least $10$ times the smallest. So they don't have the same number of digits, which contradicts obviously.

Also you can try to write periods of $\frac{1}{97},... \frac{9}{97}$. This also works...