It may be amusing to speculate on how our Sun would appear in the sky if viewed from a planet orbiting a nearby star. It’s easy enough to calculate.

Look up the equatorial coordinates of the star, and simply add exactly 180 degrees to the Declination value, and 12 hours to the Right Ascension. Then correct for the offset due to the origins of the coordinate grid on the celestial sphere.

For example, how would our sun appear from a planet of Epsilon Indi, ( distance = 3.6 parsecs, RA = 22h 03m, Dec = -56d 47′ )?

Adding 12h to the RA yields 34h 03m.
Since you cross the 0h R.A. line which is E of the star, the resulting new R.A. is 34h 03m – 24h = 10h 03m

Adding 180d to the Dec carries you to the new Dec of +56d 47′.

Since most of the naked eye stars in the night sky are bright giants at immense distances, the night sky from nearby Epsilon Indi would be very similar in appearance to the way it looks from Earth. Only a few scattered stars would appear to be shifted in location on the celestial sphere when viewed from just a few parsecs away.

The Sun’s Absolute Magnitude M (its brightness as seen from a standard distance of 10 parsecs) is 4.83. Its apparent magnitude m as it appears from any distance d is given by the magnitude -distance relation: m – M = 5 logd – 5
or
m = 5 log(3.6) – 5 + 4.83
m = 5 (0.556) – 0.17
m = 2.6

It would be an easily visible, but not particularly noteworthy, star just W of the bowl of the Big Dipper.