Doesn't Planck's formula $E=h*\nu$ tell us that all light waves have the same amplitude? Planck's constant has the units Energy-seconds. This can be interpreted as Energy divided by (cycles/second) or, equivalently Energy/Hz. The total energy for a wave of frequency, ν is h*ν. If there are 10 oscillators at that frequency then the total energy is 10*h*ν. 
Doesn't Planck's formula tell us that all waves that obey his equation have the same amplitude and differ only in frequency?
 A: If you are careful with your statement it could be true. Each oscillator is a photon. The basic equation is only for one photon, and amplitude is a classical wave concept. You are mixing classical and quantum concepts some. TThey can be mixed but with care.  
So the basic equation you stated, E = h$\nu$, is just for one photon. The amplitude is a classical term in classical electromagnetism, and does indeed represent the square root of the power. But it's for multiple photons, and it is determined by how many photons. The totals energy is the energy per photon times the number of photons, or if you want power the energy per photon times the number of photons per sec (or photon flux)
The equation P = constant x $E^2$ is valid in classical electromagnetism. In quantum electrodynamics the so called amplitude (simplistically) fluctuates. In  essence the photons collectively create the electric (and magnetic changing) fields, and so in depends on how many of them, and their phases etc. In QED (quantum electrodynamics) it is the h$\nu$ term for each photon and adding them all up, and if purely monochromatic then proportional to the number of photons. But that is only an average value, and as in all quantum theory, it obeys Heisenberg uncertainty relations: simplistically, it has a probability distribution. You measure it's average or RMS to be more exact and correct, and for one or few photons you can get different values. In the classical limit you get E. 
A: I'd be careful to accept that we know exactly what a photon is. It is true that experiments can reduce a monochromatic light source so that a "single" photon can be measured and hf is the result (within the bounds of uncertainty, of course).  As with calculations of a stronger beam, the "amplitude" calculated by QM is a probability amplitude (i.e., not a wave amplitude in the usual sense).  To make this calculation, physicists "normalize" the sum of energy possibilities using the "square" of a wave function (i.e., so that the sum of probabilities is one [1.0] - otherwise the use of statistics would be anomalous). The normalization process is a mathematical technique, not necessarily physical.  This is not to say that that the world, i.e., particles, do not physically obey laws of statistics - They do, and that fact is irrefutable.  But it is not at all physically clear that a single particle, a photon in this case, cannot have an energy of 2hf.  The normalization process would just give a calculated result (by convention essentially) as 2 photons each of energy hf. [Note that for monochromatic light, i.e., light of a single frequency, it would be improbable to have a total energy (regardless of the number of photons, or multiple energy particles) of, say, 1.5 hf or 3.6 hf, or anything but nhf (where n is an integer)]
***  Some clarifications/corrections to the original ans above....  The second response is correct, of course, that the particle wave functions are not so simple as to lead simply to nhf, and I was in error - My line of thought was one of the simplest of quantum phenomena, the photoelectric effect (as analysed by the "old QM").  That is an error, and yes, QM & QED wave functions are complex (in fact the theories cannot avoid i). To clarify what I was saying, the photoelectric effect is the easiest way. If you have a very weak beam (say 10 photons/sec) emitting light/photons at a freq. just above the threshold freq., you can detect 10 electron emissions/sec (i.e., to a high CL). If you have a high intensity beam with light at a freq. below the threshold, no electrons "should" be omitted. But this is not always the case - an election can be and does get emitted occasionally (a low probability though...). In a particle approach (e.g., QED) this is explained as two photons interacting simultaneously with one electron (actually, according to QED, they have to interact within <~10E-9 sec or so). If you take a wave approach (i.e.,not a "probability wave" or "probability amplitude") this emission is quite easily explained as an amplitude in the traditional sense. Now, even if simultaneous, we could say, "Oh, it's just two waves superimposed."... Well, that's basically an amplitude, at least in the sense I meant. Truth is that particle theories (like QED) simply do not permit photon amplitudes (except probability amplitudes). And QED is an amazingly successful theory!!! It does though, in some ways, beg the question. It is a excellent model. If you are a student I suggest you just leave it at that. Just somewhere in your mind remember that it is not reality that must conform to our models/theories, but the other way around.  Our theories of particles/photon/energy are definitely not complete. (...And that was all I meant to say at the start!)
A: 
Planck's constant has the units Energy-seconds. This can be interpreted as Energy divided by (cycles/second) or, equivalently Energy/Hz. 

So far so good. We are in the quantum mechanical framework, and a single photon has an energy E and a spin that may be pointing in its direction of motion or against it. There is no other measurable information from a single photon interacting with another particle. The frequency is only measurable through the formuka E=hnu and the nu refers to the frequency of the classical electromagnetic wave that a large number of photons build up in a superposition.
As a quantum mechanical entity, the photon is described by a wavefunction (example ),which when complex conjugate squared  gives its probability of manifesting at a point (x,y,z,t) . It is the wavefunction that carries the information of the electric and magnetic fields that will be built up  when a large number of photons make a light beam. This is because the wavefunction is a solution of a quantized  version of maxwell's equation and the classical emerges from the quantum mechanical smoothly.
I have found the following figure useful in understanding how  this happens


Left and right handed circular polarization, and their associate angular momenta.

The spin of the  individual photon is seen to be pointing at +/-1 to its direction of motion, nevertheless a circular polarization emerges from a confluence of a large number of photons, the conneciton being the spin direction of the photon.  The classical fields built up have the classical polarization.
You state:

The total energy for a wave of frequency, ν is h*ν. If there are 10 oscillators at that frequency then the total energy is 10*h*ν. 

This is not true for the photon, as the frequency is in a complex number in the wavefunction, and it is only the complex conjugate squared that is connect to the real number of energy. There is no amplitude in the singe photon. Only on the built up classical wave. To really understand how this happens one needs to study quantum electrodynamics.
A: I feel like previous answers are a bit too technical and I would like to try a simpler approach.
You are interested in number of photons and wave amplitude. One way to connect both quantities is to consider the intensity of the light beam, that is to say the power carried by the light.
From a wave perspective, Poynting theorem states that the intensity is proportional to the amplitude squared. To avoid confusions, I note $A$ the electrical field amplitude and $E_\gamma\ = h \nu$ the energy of 1 photon
$$ I \propto A^2 \Rightarrow A\propto \sqrt{I}$$
From a photon perspective, let us consider that the beam has an average photon density of $n$, each photon having the frequency $\nu$. The intensity carried by the beam is thus
$$ I \propto n\times E_{\gamma} \times c$$
So if I take two beams with the same average density of photons $n$ but two different frequency $\nu_1, \, \nu_2$, their amplitude will not be the same as
$$A_1\propto \sqrt{hnc\times \nu_1}, \,A_2\propto \sqrt{hnc\times \nu_2}$$
Contrary to your intuition, Planck formula says that waves with different frequency and identical photon number have different amplitude.
