The only such a digit is $0$, it is contained in every multiple of $10$. Let's show that no other digit is eternal for any positive integer.

Assume that some digit is eternal for integer $n$. Consider remainders of numbers
$$1, 11, 111, \dots, \underbrace{11\dots11}_{n+1}$$
modulo $n$. By the pigeonhole principle two of these remainders are equal, therefore their difference which has the form $11\ldots100\ldots0$, is a multiple of $n$. If we multiply this number by $2$ then we get a multiple of $n$ of the form $22\ldots200\ldots0$. But the only common digit for these two multiples is $0$.