No, if you assume that the size of each object is constant, O notation will swallow it and report that, in terms of the number of items to sort, Radix sort is linear in time.
The usual bound of Omega(n log n), proved using decision trees, is only applicable when your only operation is to compare two elements. Radix sort asks for more than this, and so it can assume a specific structure of its input, so it can violate the lower bound.
Depending on which operations you assume, sorting can become more or less easy. In the extreme case, if you can ask the array "Please sort yourself." as a basic operation, sorting is O(1). Radix sort assumes bitmasking as a basic operation, which falls into the "make things easier" spectrum, leading to an O(n) algorithm under the stated assumption of constant bit length (or any encoding, really, it doesn't really need to be bits).
Can you elaborate on that. It is proven that sorting based only on comparisons has a lower bound of nlog(n) comparison operations. Given that algorithms such as insertion sort use only comparison operations to probe the data, I do not see how they can break the bound.
The O(n log n) bound is a worst-case upper bound, which is a tight bound if all elements are distinct. It doesn't apply if there are guaranteed to be a sufficient number of duplicates.
For example, one might implement a variant of InsertionSort that stores a duplicate count for distinct elements, and then an insertion sort on e.g. 32-bit integers would require at most 4294967296 comparisons per insertion -- a constant factor that can be technically ignored in the complexity analysis. (I did warn you that the constant factor would become unwieldy!)
Note that this doesn't require values to be integers -- it suffices for them to be comparable with a lot of duplicates. The variant of InsertionSort described above would require O(k×n) comparisons where `n` is length of the input list and `k` is the number of distinct values, or O(n) whenever `k` is bounded by a constant (as required for radixsort to work in linear time).
That's not to say that radixsort doesn't outperform other sorting algorithms in practice -- it usually does. However, that isn't obvious from a strictly complexity theoretic point of view.