Size of Sun as seen on Pluto

New Horizons encountered Pluto when it was about 33.5 au from the Sun.

The solar diameter is $1.39times 10^{9}$ m.

An astronomical unit is $1.496times 10^{11}$ m.

The angle subtended by the Sun at Pluto (in radians) is the ratio of the diameter of the Sun to its distance from Pluto. Converting to degrees, minutes, seconds, gives 0.01589 degrees, or 0.95 arcminutes.
The atmosphere of Pluto is way too thin (much thinner than the Earth's) to make any significant difference to this at all.

The diameter of Pluto is $2.37times 10^6$ m. In order for Pluto and the Sun to subtend a similar angle in an image, you just need to move far enough away from Pluto to make the same angle.

Thus

$$D_P/R_P simeq (D_P + D_{PS})/R_S,$$
where $D_P$ would be the distance from New Horizons to Pluto, $D_{PS}$ the distance from Pluto to the Sun, $R_P$ the radius of Pluto and $R_S$ the radius of the Sun. Because I will show that $D_P ll D_{PS}$ we can say that
$$D_P simeq frac{R_P}{R_S} D_{PS} simeq 0.057 au$$ .

New Horizons is travelling at about 14.5 km/s. So a simple bit of Maths tells us it would takes about 6.5 days to get into this position.

The picture I think you are talking about was taken on July 15th 2015 only 7 hours after the flyby. The Sun and Pluto do not share the same angular size in this photo. You are not seeing the Sun at all in this picture. It is eclipsed by Pluto completely. All that is being seen is sunlight refracted (or more likely scattered) through Pluto's atmosphere. The Sun acts more-or-less like a point source of light at more-or-less infinite distance.

The angle that the light needs to deviate in order to be seen through the atmosphere is about $0.19$ degrees (the half-angle subtended by Pluto at the Sun [negligible] plus the angle subtended by Pluto at New Horizons after 7 hours). If we were to model the atmosphere as a glass prism. The deviation angle is $delta simeq (n-1) alpha$, where $n$ is the refractive index and $alpha$ is the opening angle of the prism. If you let $alpha$ be quite large (let's say 1 radian) to simulate rays coming through the atmosphere, then $n = 1 + delta$, where $delta$ is in radians. So $nsimeq 1.003$. NB: This assumes there is no scattering in Pluto's atmosphere, but I suspect that is not true and that my estimate of the refractive index is a large overestimate. In fact I'm sure it is an overestimate because this is larger than the refractive index of air on Earth. i.e. I am sure that what we are seeing is scattered light, not refracted light.

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