Numbers with Rotational Symmetry Given an integer, output a truthy value if it is the same upside-down (rotated 180°) or a falsy value otherwise.

0, 1, and 8 have rotational symmetry. 6 becomes 9 and vice versa.

Sequence of numbers producing truthy results: OEIS A000787

0, 1, 8, 11, 69, 88, 96, 101, 111, 181, 609, 619, 689, 808, 818, 888, 906, 916, 986, 1001, 1111, 1691, 1881, 1961, 6009, 6119, 6699, 6889, 6969, 8008, 8118, 8698, 8888, 8968, 9006, 9116, 9696, 9886, 9966, 10001, 10101, 10801, 11011, 11111, 11811, 16091, ...


This question is inspired by my own reputation at the time of posting: 6009.
Python 2, 50 bytes

lambda n:`n`==`map('01xxxx9x86'.find,`n`)`[-2::-3]

The method '01xxxx9x86'.find takes a digit character to it's upside-down number, with any unflippable digit giving -1. This function is mapped to the reversed number string, producing a list of digits.
This is converted to a string with the [1::3] trick, except it's reversed instead by doing [-2::-3] (thanks to Dennis for this, saving 4 bytes),  and compared to the original number string. Any -1's from unflippable digits will misalign the conversion, making it fail.

56 bytes:
lambda n:`n`[::-1]==`n`.translate('01xxxx9x86______'*16)

Checks if the number string reversed is the same as it with the upside-down replacements. Digits that can't be flipped are replaced with 'x' to always give the wrong answer.
The replacement is done with translate on a string of 256 chars, replacing the corresponding ASCII values. Only the 10 values 48 to 57 matter, but I padded to length 16 to make the total length be 256. I wonder if there's a shorter way.
Some other approaches (lengths 59, 60, 60):
lambda n:set(zip(`n`,`n`[::-1]))&lt;=set(zip('01896','01869'))
r=lambda s:set(zip(s,s[::-1]));lambda n:r(`n`)&lt;=r('9018106')
lambda n:all(a+b in'001188969'for a,b in zip(`n`,`n`[::-1]))