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.
Everything that passes the event horizon of a black hole falls into the singularity, including photons. That's why it's a singularity.
There is a particular radius outside the event horizon where a photon will orbit, but the orbit is unstable- if the photon gets perturbed a little closer or the black hole's mass increases at all, it will fall in, and if the black hole's mass decreases due to Hawking radiation or the photon gets perturbed away from the black hole, it will escape.
Best Answer
The photon sphere is of theoretical interest only- the photon orbits are unstable [a boulder at the top of a hill, instead of the bottom of a valley], so in the real world there isn't going to be some huge herd of photons to detect or crash through or anything like that.
In the same vein, if you fall through it, you fall through it. Nothing special happens to you. Other than the fact that you are almost certainly going to have a bad experience very soon, if you are not already.
No. The photon is captured if its impact parameter (that is, the distance "of closest approach" to the black hole) is any less than the radius of the photon sphere. That's the definition of the photon sphere. In other words, any photon that approaches from distant space will be captured if its path goes inside the photon sphere at all. However, a photon emitted on a radial trajectory from an object between the event horizon and the photon sphere can escape. A photon emitted by an object inside the event horizon has nowhere to go but towards the singularity.
Neutron stars do not have a photon sphere. The Schwarzschild radius of a neutron star is about 3 km, which means you have to compress it within that size to create a black hole. This means that the photon sphere radius would be about 4.5 km, so to have a photon sphere you would need to cram the neutron star smaller than that. [Warning: Do not try this at home. You will create a supernova, with a black hole remnant.] Neutron stars are actually more like 10-15 km in radius, so they do not have strong enough gravity to do any of that.