Earlier quoted context omitted.
No, radix sort is worst case O(n * k). In many common cases, k ~= log(n). IN certain specific cases, k < log(n), and specifically for cases where you have a very large n, but a bounded number of values (say, you're sorting 10 billion 4-bit ints), k can be considered a constant. But that is by no means generally true.
In most cases k << n. For 64-bit integers, byte wise radix sort k is 8, which is less than log n whenever n is more than 256. So, radix sort is typically much faster than an O(n log n) sort of your data support it. It just isn’t as widely used because it is not as general as a comparison based sort.
In the worst case, every element falls into a single bucket, at which point your best best is to do a bit wise radix sort over the low 8 bits.
This ends up being equivalent to k=log(n).