How does the view of an observer near an event horizon change as a function of the observer's position, velocity, and acceleration? I see that this topic has been well covered from the perspective of a) a stationary observer dangling on the end of a rope and and b) an observer free-falling from infinity.  What I'm still unsure about are the roles of velocity vs. acceleration. Is it all about velocity?  Does an observer who is stationary near the horizon and only just starting to freefall see the same as a)?  Also, for an observer who is descending at close to escape velocity but then fires his/her retrorockets to cancel the acceleration and achieve constant velocity, does he/she still see the same as b)?
 A: What you see is what lies on your past light cone. The past light cone only depends on spacetime position, not velocity or higher derivatives.
What you see also depends on your velocity in the sense that it is "distorted" in different ways by Doppler shift and aberration. But given a (infinite-precision, omnidirectional) photograph of what someone at a particular location with a particular velocity sees, you can derive what someone at the same location with any other velocity will see just from that picture, without needing to know anything about the 4D spacetime it was originally derived from.
Acceleration doesn't affect what an idealized camera/eye sees.

Does an observer who is stationary near the horizon and only just starting to freefall see the same as a)? Also, for an observer who is descending at close to escape velocity but then fires his/her retrorockets to cancel the acceleration and achieve constant velocity, does he/she still see the same as b)?

Yes to both questions, and also, if you ignore Doppler shift and aberration, all four of them see the same thing when they're at the same spacetime location even if they don't match their velocities.
