Don't mangle pd.NaT and np.nan in dp.unqiue()

[realead]

Code Sample, a copy-pastable example if possible

import pandas as pd
pd.unique([np.nan, pd.NaT])  # returns [nan]

Problem description

This result is a little bit inconsistent, because:

import pandas.core.algorithms as algos
algos.isin([np.nan], [pd.NaT])  

returns False, i.e. np.nan and pd.NaT are not the same.

Expected Output

pd.unique([np.nan, pd.NaT]) should return [np.nan, pd.NaT].

Output of pd.show_versions()

INSTALLED VERSIONS

commit: None
python: 3.6.2.final.0
python-bits: 64
OS: Linux
OS-release: 4.4.0-53-generic
machine: x86_64
processor: x86_64
byteorder: little
LC_ALL: None
LANG: en_US.UTF-8
LOCALE: en_US.UTF-8

pandas: 0.22.0
pytest: 3.2.1
pip: 10.0.1
setuptools: 36.5.0.post20170921
Cython: 0.28.3
numpy: 1.13.1
scipy: 1.1.0
pyarrow: None
xarray: None
IPython: 6.1.0
sphinx: 1.6.3
patsy: 0.4.1
dateutil: 2.6.1
pytz: 2017.2
blosc: None
bottleneck: 1.2.1
tables: 3.4.2
numexpr: 2.6.2
feather: None
matplotlib: 2.0.2
openpyxl: 2.4.8
xlrd: 1.1.0
xlwt: 1.3.0
xlsxwriter: 0.9.8
lxml: 3.8.0
bs4: 4.6.0
html5lib: 0.9999999
sqlalchemy: 1.1.13
pymysql: None
psycopg2: None
jinja2: 2.9.6
s3fs: 0.1.3
fastparquet: None
pandas_gbq: None
pandas_datareader: None