On Digts of Cubes and Squares This is a pretty nice site I found, which contains special properties of all numbers between $1$ and $9999$, with the exception of few.
The number I noticed was $69$. All digits appear exactly once in  $69^2=4761$, and $69^3=328509$.
However, is the following true?

Are there an infinite number of $i$ for which there exists such an interger $n$ that $n^2$ and $n^3$ contain all digits exactly $i$ times?

Since a square or cube does not normally have a lot of same numbers, it would seem that the answer is false. However, I was not able to prove or disprove the above.
It is clear that if such a $n$ existed for $i$, $n \equiv 0 \pmod 3$ or $n \equiv 8 \pmod 9$ through modular inspection.
Any help would be appreciated.
 A: Here is an extremely handwavy argument suggesting that there should be many examples.  To have a chance this works we need $n^2$ and $n^3$ to have exactly $10i$ digits between them.  This requires that $4.64 \cdot 10^{2i-1} \lt n \lt 10^{2i}$, where the $4.64$ is the cube root of $100$.  This forces $n^2$ to have $4i$ digits and $n^3$ to have $6i$ digits.  We now make the wild assumption that for large $i$ the digits are randomly selected and ask the probability that we get exactly $i$ of each digit.  As there are $10i$ digits, we can choose the places for zeros in $10i \choose i$ ways.  Having done that, we can choose the places for ones in $9i \choose i$ ways.  The chance we get exactly the same number of each digit is then $$p(i)=\frac {{10i \choose i}{9i \choose i}{8i \choose i}\dots {2i \choose i}}{10^{10i}}=\frac {(10i)!}{(i!)^{10}10^{10i}}$$  If we use Stirling's approximation we get $$p=\frac{(10i)^{10i}e^{10i}\sqrt{2\pi (10i)}}{i^{10i}10^{10i}e^{10i}(\sqrt{2\pi i})^{10}}=\frac{\sqrt{10}}{(2\pi i)^{9/2}}$$  For $i=4$ this is about $1.6 \cdot 10^{-6}$ while the exact answer is about $1.3 \cdot 10^{-6}$.  As there are over $5 \cdot 10^7$ numbers in our range we should expect tens of examples.  This is a number that would be easy to search by computer.  As $i$ gets larger the expected number of examples per decade will grow rapidly.  The $i^{9/2}$ divisor gets swamped by the $10^{2i}$ multiplier from the range of $n$.
