I don't see a way to do this using the
zip command, but it's easy in python.
Note that the zip file format specification, section 188.8.131.52, says that the pathname cannot start with a '/', so I can't help with that part.
The python zipfile module will let you override a file's pathname when you add it to the zip archive; just pass the desired name as the optional 2nd argument to
ZipFile.write(filename[, arcname[, compress_type]])
Write the file named filename to the archive, giving it the archive name arcname (by default, this will be the same as filename, but without a drive letter and with leading path separators removed). If given, compress_type overrides the value given for the compression parameter to the constructor for the new entry.
Note: Archive names should be relative to the archive root, that is, they should not start with a path separator.
Here's an example:
$ touch 1 2 3
Python 2.7.6 (default, Mar 22 2014, 22:59:56)
>>> import zipfile
>>> with zipfile.ZipFile('bundle.zip', 'w') as bundle:
... bundle.write('1', '/bin/1')
... bundle.write('2', '/sbin/2')
... bundle.write('3', '/usr/bin/3')
$ unzip -l bundle
Length Date Time Name
--------- ---------- ----- ----
0 2014-08-11 13:00 bin/1
0 2014-08-11 13:00 sbin/2
0 2014-08-11 13:00 usr/bin/3
0 3 files
zipfile.write will remove any leading '/' from the pathname, to conform with the standard.