You don't have to orbit, you can just use a rocket to stay put. All observers that can communicate with infinity for all time agree about the infalling object. It gets frozen and redshifted at the horizon.
EDIT: in response to question
The issue of two objects falling in one after the other is adressed in this question: How does this thought experiment not rule out black holes? . The answers there are all wrong, except mine (this is not an arrogant statement, but a statement of an unfortunate fact).
When you are near a black hole, in order to stay in place, you need to accelerate away from the black hole. If you don't, you fall in. Whenever you accelerate, even in empty Minkowksi space, you see an acceleration event horizon behind you in the direction opposite your acceleration vector. This horizon is a big black wall that follows you around, and you can attribute the various effects you see in the accelerating frame, like the uniform gravitational field and the Unruh radiation, to this black-wall horizon that follows you around.
When you are very near a black hole, staying put, your acceleration horizon coincides with the event horizon, and there is no way to tell them apart locally. This is the equivalence principle, in the form that it takes in the region by the horizon where there is no significant curvature.
The near-horizon Rindler form of the metric allows you to translate any experiment you can do in the frame near a black hole to a flat space with an accelerating observer. So if you measure the local Hawking temperature, it coincides with the Unruh temperature. If you see an object fall and get redshifted, you would see the same thing in empty space, when accelerating.
The point is that the acceleration you need to avoid falling in is only determined globally, from the condition that you stay in communication with infinity. If you stop accelerating so that you see the particle cross the horizon, the moment you see the particle past the horizon, you've crossed yourself.
You need to be a lot more careful when you use the phrase red-shift, due to how frequency is measured in general relativity. Roughly speaking, a photon is characterised by its wave vector $k$, which is a light-like four vector. The frequency measured by an observer is $g(\tau,k)$ where $g$ is the metric tensor, and $\tau$ is the unit vector tangent to the observer's world-line.
A little bit of basic Lorentzian geometry tells you that for any given photon $k$, instantaneously there can be observers seeing that photon with arbitrary high and arbitrarily low frequency.
So: start with your space-time. Fix a point on the event horizon. Fix a photon passing through that space-time event. For any frequency you want to see, you can choose a time-like vector at that space-time event that realises that frequency. Now, since the vector is time-like and the event horizon is null, the geodesic generated by that vector must start from outside the event horizon and crosses inside. Being a geodesic, it represents a free fall. So the conclusion is:
For any frequency you want to see, you can find a free falling observer starting outside of the black hole, such that it crosses the event horizon at the given space-time event and observes the frequency you want him to see.
So you ask, what is this whole business about gravitational red-shift of Schwarzschild black holes? I wrote a longer blog post on this topic some time ago and I won't be as detailed here. But the point is that on the Schwarzschild black holes (and in general, on any spherically symmetric solution of the Einstein's equations), one can break the freedom given by local Lorentz invariance by using the global geometry.
On Schwarzschild we have that the solution is stationary. Hence we can use the time-like Killing vector field* for the time-translation symmetry as a "global ruler" with respect to which to measure the frequency of photons. This is what it is meant by "gravitational redshift" in most textbooks on general relativity (see, e.g. Wald). Note that since we fixed a background ruler, the frequency that is being talked about is different from the frequency "as seen by an arbitrary infalling observer".
(There is another sense in which redshift is often talked about, which involves two infalling observers, one "departing first" with the second "to follow". In this case you again need the time-translation symmetry to make sense of the statement that the second observer "departed from the same spatial point as the first observer, but at a later time.)
It turns out, for general spherically symmetric solutions, there is this thing called a Kodama vector field, which happens to coincide with the Killing vector field on Schwarzschild. Outside of the event horizon, the Kodama vector field is time-like, and hence can be used as a substitute for the global ruler with respect to which to measure red-shift, when the space-time is assumed to be spherically symmetric, but not necessarily stationary. Again, this notion of redshift is observer independent. And it has played important roles (though sometimes manifesting in ways that are not immediately apparently related to red-shift, through choices of coordinates and what-not) in the study of dynamical, spherically symmetric gravitational collapse in the mathematical physics literature.
To summarise:
If you just compare the frequency of light measured (a) at its emission at the surface of the star in the rest frame associated to the collapse and (b) by an arbitrary free-falling observer, you can get basically any values you want. (Basically because the Doppler effect depends on the velocity of the observer, and you can change that to anything you like by choosing appropriate initial data for the free fall.)
One last comment about your last question:
You asked about what happens in the interior of the black hole. Again, any frequency can be realised by time-like observers locally. The question then boils down to whether you can construct such time-like observers to have come from free fall starting outside the black hole. By basic causality considerations, if you start with a time-like vector at a space-time event inside the black hole formed from gravitational collapse, going backwards along the time-like geodesic generated by the vector you will either hit the surface of your star, or exit the black hole. Though precisely how the two are divided depends on the precise nature of the gravitational collapse.
I should add that if you use the "global ruler" point of view, arguments have been put forth that analogous to how one expects red shift near the event horizon, one should also expect blue shifts near any Cauchy horizon that should exist. This has been demonstrated (mathematically) in the Reissner-Nordstrom (and similar) black holes. But as even the red-shift can sometimes run into problems (extreme charged black holes), one should not expect the statement about blue shifts near the Cauchy horizons to be true for all space-times.
Best Answer
I believe I understand what you are asking and the answer is that the observer does not see an accurate representation of what happens near the event horizon. In the first graph the line would curve up to infinity at the end because as the object fell into the hole the light would be slowed to a stop at the event horizon. As the object begins it's approach to the Black Hole at point A, there is not sufficient gravity to affect the speed of light emitted by the object. As the object gets very close to the Event Horizon and the gravity of the Black Hole has a considerable effect on the speed of light, the object appears to slow down from the perspective of the observer, but not relative to itself. Finally, immediately before the Event Horizon is reached the emitted light from the object is brought to an infinitely low speed and appears to the observer to stall.
An interesting follow-up question is would the object get dimmer as it approached the Event Horizon because the flux of photons was reduced?