Concerning ourselves only with searching for automorphic numbers $n$ in bases $1 < b < 17$ (binary to hexadecimal) and the ranges given by the iterator $0 < i < b^3 - 1$ and limiting to $m = 1$ we find the following results:

First, it is obvious that 1 is a 1-automorphic number regardless of the base.

For the range and limit specified, there are no other 1-automorphic numbers in binary through quinary, bases 7 through 9, 11, 13 and 16.

In base 6, there are 1, 3, 4, 9, 28, 81, 136, and it is easy to verify that ${3_6}^2 = 13_6$ ${4_6}^2 = 24_6$ ${13_6}^2 = 213_6$ ${44_6}^2 = 3344_6$ etc.

In base 10, these ought to look familiar: 1, 5, 6, 25, 76, 376, 625.

Duodecimal: 1, 4, 9, 64, 81, 513, 1216. Noticing that 4 also appears in the list for base 6, we might wonder if 4 is always 1-automorphic when $6|b$ The question is moot because the next multiple of 6 is $18 > 4^2$ thus in base 18 and any other higher bases, the square of 4 is also a 1-digit number.

Base 14: 1, 7, 8, 49, 148, 344, 2401.

Base 15: 1, 6, 10, 100, 126, 1000, 2376. Base 15 is the smallest odd base $b$ to have 1-automorphic numbers in the range specified. This should not be taken to imply that it is the smallest odd base to have automorphic numbers at all.