float64 does not respect __rmul__() protocol and does unexpected things instead.
I didn't test, but it might be similar for the other operators.

Reproducing code example:

import numpy as np
x = np.float64(2)
x * (1,)

raises a TypeError (expected), while __mul__((1,)) correctly returns NotImplemented.

import numpy as np
class Test(tuple):
    def __rmul__(self, other):
        return None
x = np.float64(2)
x * Test((1,))

returns array([2]), what is not correct (None is the expected result).
On the contrary, 2 * Test((1,)) correctly returns None.

NumPy/Python version information:

python: 3.8.2
numpy: 1.19.2