How exactly are the degrees of freedom seen by a falling into a black hole observer related to the ones seen by a staying outside observer? This is some kind of a follow up of this nicely to the point answer to a provocative (but nevertheless upvoted!) question, about the legitimacy of black hole physics. The  answer mentions, that the interior and exterior of a black hole are not completely decoupled, and about this exact relationship I'd like to learn some more not just hand waving details. 
Of course I know the complementarity principle, which says that the perspective of what happens near the event horizon of a black hole an infalling observer who does not note anything special when crossing the event horizon and the description of the physics near the horizon by a staying outside observer who sees the infalling observer getting frozen on the horizon, are both equally legitimate descriptions of the same physics.
Is there some kind of a map, or exact dictionary by which one can transform back and forth between these two descriptions of the same physics, that describes the complementary principle for black holes mathematically?
An aside: I always thought that the holographic principle is soemething different from the complementarity principle in this context, so this should not be the trivial answer to my question (?) ... Maybe I have managed to confuse myself now :-)
Clarification after some comments
I am not just up to a coordinate transformation between spacetime inside and outside the even horizon, but I'd like to learn about how the degrees of freedom used by an infalling observer and the ones used by a staying outside observer to describe the same physics transform into each other.
 A: I share the same big interest about your question and I have not the answer, but I have one idea: 
The idea is to study a simpler example, that is a non-uniform accelerating observer in special relativity.
For instance, imagine a 100-meter race in a stadium. Typically,  it takes 10 seconds for the runners to finish the race. 
The people in the stadium (as observers) see the runner crossing the finish line.
We may call $z$ and $t$ the coordinates as seen by a stadium observer (S), so as the runner crosses the finish line $z = 0$ at $t = 0$. 
Now, we can choose an moving observer (M), which has a variable speed $v(t)$ relatively to the stadium observer. I think that one can choose $v(t)$ such as the observer M never sees the runner crossing the finish line. 
You need simply: $$\int_{-10}^0 \frac{dt}{\sqrt{1 - \frac{v^2(t)}{c^2}}} = + \infty$$
(I suppose here that we consider that the speed of the runner is neglectible relatively to the speed of light)
So, it is a kind of horizon, it is a planar horizon at $z = 0$, instead of a spherical horizon in the Schwartzschild GR problem, but the logic is the same.
Now, by knowing $v(t)$, you know the transformation law between $dz'$, $dt'$ (M coordinates) and $dz$, $dt$ (S coordinates)
From this, it would be possible, in principle, to get the transformation of fields, for instance, for a scalar field, we would have (with $x'=x$ and $y'=y$): 
$$\phi'(x', y', z', t') = \phi(x, y, z, t)$$
For instance, it would be possible to modelize the runner as a cylindric or cubic wave-packet $\phi$, and we should be able to prove that we have : 
$$\phi'(x', y', z', t') = 0 ~ for ~ z > 0$$
