> essentially fixed in the sky

Let's start with "fixed in the sky" and qualify your frame of reference as the field of distant stars, or the celestial sphere. The common coordinate system is right ascension (RA) and declination (dec).

The GP question was about the Earth's rotation, which would be in terms of azimuth and altitude, and that question's been asked and answered. The key terms there: "equatorial mount" and "clock drive".

The comet C/2025 R3 (PanSTARRS) is in fact near its highest velocity (with reference to the Sun especially), being near perihelion while this photograph was taken. The comet is swinging around the Sun, and it was about 0.49 AU from Earth at the time of the photograph.

https://en.wikipedia.org/wiki/C/2025_R3_(PanSTARRS)

I chose an approximate time on April 27: in 10 minutes of wall-clock time, with the J2000 epoch, the comet's apparent motion is from RA 02h 49m 07.1s, dec +06° 02' 56.5" to RA 02h 49m 15.4s, dec +06° 02' 13.3"

That is a distance of 2' 11.13" across the celestial sphere. For reference, Venus is 11.6" wide in the sky as we see it this week.

24 hours later, we find it at RA 3h 08m 44.1s, dec +04° 19' 27.8". Its apparent motion was 5° 10' 46.02", which is approximately the width of your three middle fingers held together, at arm's length.

So, "fixed in the sky" is not a scientifically useful description of astronomical objects: we need to put that in terms of at least one frame of reference, and "apparent motion" which is how an observer perceives it.

https://soho.nascom.nasa.gov/data/LATEST/latest-lascoC3.html (grab this today; scroll between 4/23 and 4/27)

Given that the moon is about 0.5 degrees in diameter from Earth, shouldn't we expect to see the stars and comet much more blurred than they are here though? Or the ground if it's stabilized against the rotation?

40 degrees around Earth (central angle)

but it increases to much more when you are much closer to the arc