pydrake math: .multiply shouldn't lose shape information

[EricCousineau-TRI]

@kunimatsu-tri ran into the following issue:

import numpy as np
from pydrake.math import RigidTransform

a = np.array([0., 0., 0.])
b = a.reshape((3, 1))
c = np.array([[1., 2., 3.], [4., 5., 6]]).T
X = RigidTransform()

for v in [a, b, c]:
    mul_shape = (X @ v).shape
    print(f"{v.shape} => {mul_shape}")

Output:

(3,) => (3,)
(3, 1) => (3,)
(3, 2) => (3, 2)

Note how the case with b loses shape information, from (3, 1) to (3,).

From docs:
https://drake.mit.edu/pydrake/pydrake.math.html#pydrake.math.RigidTransform_.RigidTransform_[float].multiply

Specifically, we're looking at the interplay btw these two overloads:

2. multiply(self: pydrake.math.RigidTransform_[float], p_BoQ_B: numpy.ndarray[numpy.float64[3, 1]]) -> numpy.ndarray[numpy.float64[3, 1]]

3. multiply(self: pydrake.math.RigidTransform_[float], p_BoQ_B: numpy.ndarray[numpy.float64[3, n]]) -> numpy.ndarray[numpy.float64[3, n]]

In pybind11, float64[3, 1] gets intercepted as a vector (overload 2), and when it says it returns float64[3, 1], it really means it returns float64[3].

[EricCousineau-TRI]

Relates this section:
https://pybind11.readthedocs.io/en/stable/advanced/cast/eigen.html#vectors-versus-column-row-matrices