Here's a solution, which, although not 100% vectorized, (per my benchmarks) takes about half the time as your does (using your sample data). The performance improvement probably becomes more drastic with more data:

maximums = [a.max() for a in np.split(values, np.arange(1, ids.shape[0])[(np.diff(ids) != 0)])]

Output:

>>> maximums
[5, 3, 6, 8, 8]

In trying to visualize the problem:

In [82]: [np.where(ids==id) for id in uniq_ids]
Out[82]: 
[(array([0]),),
 (array([1]),),
 (array([2, 3, 4, 5]),),
 (array([6]),),
 (array([7, 8, 9]),)]

unique can also return:

In [83]: np.unique(ids, return_inverse=True)
Out[83]: (array([0, 1, 3, 5, 6]), array([0, 1, 2, 2, 2, 2, 3, 4, 4, 4]))

Which is a variante on what richardec produced:

In [88]: [a for a in np.split(ids, np.arange(1, ids.shape[0])[(np.diff(ids) !=
    ...: 0)])]
Out[88]: [array([0]), array([1]), array([3, 3, 3, 3]), array([5]), array([6, 6, 6])]

That inverse is also produced by doing where on all == at once:

In [90]: ids[:,None] == uniq_ids
Out[90]: 
array([[ True, False, False, False, False],
       [False,  True, False, False, False],
       [False, False,  True, False, False],
       [False, False,  True, False, False],
       [False, False,  True, False, False],
       [False, False,  True, False, False],
       [False, False, False,  True, False],
       [False, False, False, False,  True],
       [False, False, False, False,  True],
       [False, False, False, False,  True]])
In [91]: np.nonzero(ids[:,None] == uniq_ids)
Out[91]: (array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]), array([0, 1, 2, 2, 2, 2, 3, 4, 4, 4]))

I'm still thinking through this ...

EDIT: I'll leave this up as an example of why we can't always use np.vectorize() to make everything magically faster:

One solution is to use numpy's vectorize function:

import numpy as np

values = np.array([5, 3, 2, 6, 3, 4, 8, 2, 4, 8])
ids = np.array([0, 1, 3, 3, 3, 3, 5, 6, 6, 6])

def my_func(id):
    return np.max(values[np.where(ids==id)])

vector_func = np.vectorize(my_func)
maximums = vector_func(np.unique(ids))

which returns

array([5, 3, 6, 8, 8])

But as for speed, your version has about the same performance when we use

values = np.array([random.randint(1, 100) for i in range(1000000)])
ids = []
for i in range(100000):
    r = random.randint(1, 4)
    if r == 3:
        for x in range(3):
            ids.append(i)
    elif r == 2:
        for x in range(4):
            ids.append(i)
    else:
        ids.append(i)
ids = np.array(ids)

It's about 12 seconds per execution.

With pandas:

import pandas as pd

def with_pandas(ids, vals):
    df = pd.DataFrame({'ids': ids, 'vals': values})
    return df.groupby('ids')['vals'].max().to_numpy()

Timing:

import numpy as np

values = np.random.randint(10000, size=10000)
ids = np.random.randint(100, size=10000)

%timeit with_pandas(ids, values)
692 µs ± 21.1 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

np.lexsort sorts by multiple columns. However, this is not compulsory. You can sort ids first and then choose maximum item of each divided group using numpy.maximum.reduceat

def mathfux(values, ids, return_groups=False):
    argidx = np.argsort(ids) #70% time
    ids_sort, values_sort = ids[argidx], values[argidx] #4% time
    div_points = np.r_[0, np.flatnonzero(np.diff(ids_sort)) + 1] #11% time (the most part for np.flatnonzero)
    if return_groups: 
        return ids[div_points], np.maximum.reduceat(values_sort, div_points)
    else: return np.maximum.reduceat(values_sort, div_points)

mathfux(values, ids, return_groups=True)
>>> (array([0, 1, 3, 5, 6]), array([5, 3, 6, 8, 8]))
mathfux(values, ids)
>>> mathfux(values, ids)
array([5, 3, 6, 8, 8])

Usually, some parts of numpy codes could be optimised further in numba. Note that np.argsort is a bottleneck in majority of groupby problems which can't be replaced by any other method. It is unlikely to be improved soon in numba or numpy. So you are reaching an optimal performance here and can't do much in further optimisations.