BUG: groupby.first() corrupts timezone
#### Code Sample, a copy-pastable example if possible

```
import pandas as pd
df = pd.DataFrame([{'ts': pd.Timestamp('2016-01-01', tz='America/Los_Angeles'), 'a': 1}])
df.groupby('a').first()
```

The output is

```
                         ts
a                          
1 2016-01-01 08:00:00-08:00
```
#### Expected Output

```
                         ts
a                          
1 2016-01-01 00:00:00-08:00
```

Note that the issue is with `first()` and not `groupby`:

```
for _, group in df.groupby('a'):
    print(group.ix[0])
```

gives the correct output of 

```
a                             1
ts    2016-01-01 00:00:00-08:00
Name: 0, dtype: object
```
#### output of `pd.show_versions()`

```
pd.show_versions()

INSTALLED VERSIONS
------------------
commit: None
python: 2.7.6.final.0
python-bits: 64
OS: Linux
OS-release: 3.16.0-60-generic
machine: x86_64
processor: x86_64
byteorder: little
LC_ALL: None
LANG: en_US.UTF-8

pandas: 0.18.0
nose: 1.3.1
pip: 1.5.4
setuptools: 3.3
Cython: None
numpy: 1.10.4
scipy: 0.13.3
statsmodels: 0.5.0
xarray: None
IPython: 4.0.0
sphinx: None
patsy: 0.2.1
dateutil: 2.5.1
pytz: 2016.3
blosc: None
bottleneck: None
tables: 3.1.1
numexpr: 2.2.2
matplotlib: 1.3.1
openpyxl: 1.7.0
xlrd: 0.9.2
xlwt: 0.7.5
xlsxwriter: None
lxml: 3.3.3
bs4: 4.2.1
html5lib: 0.999
httplib2: 0.8
apiclient: None
sqlalchemy: None
pymysql: None
psycopg2: None
jinja2: 2.8
boto: None
```
