If I take two plane EM waves travelling in opposite direction e.g. $E = E_0 \sin(kx-\omega t)$ and $E=E_0 \sin (kx + \omega t)$, they sum to give a standing wave with a time-averaged Poynting vector of zero.
If I use the appropriate special relativistic transformations to derive how these fields appear to an observer travelling at $v$ along the x-axis, I find that one E-field is diminished, one is boosted, whilst at the same time one wave is blue-shifted and the other red-shifted. These waves do not sum to give a standing wave in the moving frame of reference and have a non-zero time-averaged Poynting vector.
So, is the phenomenon of a standing wave dependent on the frame of reference of the observer?
Please note: That I have tried transforming the fields and summing them, but could not make sense of the result in terms of a travelling wave or a standing wave – hence the question. To get the bounty I'm looking for a form for $E^{\prime}$ or a decomposition of $E^{\prime}$ (that isn't just the sum of the transformed waves!) that makes clear the nature of the summed waves in the moving frame.
Best Answer
Yes, the concept of a standing wave is frame dependent, and a standing wave selects a specific frame just like the concept of a rest frame.
My initial answer was similar to BenCrowell's, and I made the same oversimplification mistake that his initial answer does. The are no spatial coordinates for the standing wave (even in its rest frame) in which the fields are a constant $\mathbf{E}=\mathbf{B}=0$. The existence of such would simplify the mental image of this scenario greatly, but unfortunately that is not the case.
However in the rest frame there are spatial planes in which the electric field is always zero, and other planes in which the magnetic field are always zero. There are even discrete times at which on spatial planes, both $\mathbf{E}$ and $\mathbf{B}$ are zero. Since these do not exist at all times, it becomes a bit difficult to discuss them 'moving', but a similar idea does appear.
In the original frame, away from a node the time averaged Poynting vector is zero, but the energy density is not zero. When transforming to the new frame, the components of the momentum and energy density will mix up (technically the stress energy tensor). So it should make conceptual sense that after the transformation the time averaged momentum density is no longer zero (also seen conceptually via your red/blue shifting argument).
So, does this count as a "standing wave"?
No. There are no longer constant node planes of $\mathbf{E}=0$ or $\mathbf{B}=0$, and there is now non-zero time averaged momentum density (so the stress energy tensor is that of something "moving" not "standing").
Calculation details:
Consider a standing wave formed from two plane waves travelling in the +/- x direction, linearly polarized in the z direction: $$\mathbf{E} = \mathbf{\hat{z}} E_0\left[\sin(kx - \omega t) + \sin(-kx - \omega t)\right] $$ $$\mathbf{B} = \mathbf{\hat{y}} B_0\left[-\sin(kx - \omega t) + \sin(-kx - \omega t)\right] $$
For a single plane wave, the nodes where $\mathbf{E}=\mathbf{B}=0$ are moving planes in space (so contiguous volumes in spacetime). For the above sum of waves, the nodes are still planes in space, but only show up at discrete times.
Looking at where the electric field is zero: $$\begin{align*} 0 &= \left[\sin(kx - \omega t) + \sin(-kx - \omega t)\right] \\ &= \sin(kx)\cos(\omega t) - \cos(kx)\sin(\omega t) - \sin(kx)\cos(\omega t) - \cos(kx)\sin(\omega t) \\ &= -2 \cos(kx)\sin(\omega t) \end{align*} $$ Similarly for the magnetic we will find: $$0 = -2 \sin(kx)\cos(\omega t)$$ which is similar, just shifted by a quarter wavelength in space and quarter period in time.
Focusing on the nodes in the electric field first, these are planes located at: $$ kx = \pi \left(n+\frac{1}{2}\right) \quad \rightarrow \quad x = a_n = \frac{\pi}{k}\left(n+\frac{1}{2}\right) $$ with $n$ any integer, for any value of $y,z,t$ (so a constant plane in space). The electric field will also be zero everywhere when $\omega t = \pi m$ with $m$ any integer.
For the magnetic field, there are planes located at: $$ kx = \pi n \quad \rightarrow \quad x = b_n = \frac{\pi n}{k} $$ with $n$ any integer, for any value of $y,z,t$ (again a constant plane in space). The magnetic field will also be zero everywhere when $\omega t = \pi (m+1/2)$ with $m$ any integer.
We therefore see spatial planes that show up with $\mathbf{E}=\mathbf{B}=0$ at $x=a_n,t=\pi (m+1/2)$ and at $x=b_n,t=\pi m$.
Let's define the new coordinates as a Lorentz transformation for the original coordinate system: $$\begin{align*} ct' &= \gamma(ct - \beta x) \\ x' &= \gamma(x - \beta ct) \\ y' &= y \\ z' &= z \end{align*}$$ with $\beta=\frac{v}{c},\gamma=[1-\beta^2]^{1/2}$.
The transformation laws written out for the fields gives the fields in this new frame as: $$\begin{align} & \mathbf {{E}_{\parallel}}' = \mathbf {{E}_{\parallel}}\\ & \mathbf {{B}_{\parallel}}' = \mathbf {{B}_{\parallel}}\\ & \mathbf {{E}_{\bot}}'= \gamma \left( \mathbf {E}_{\bot} + \mathbf{ v} \times \mathbf {B} \right) \\ & \mathbf {{B}_{\bot}}'= \gamma \left( \mathbf {B}_{\bot} -\frac{1}{c^2} \mathbf{ v} \times \mathbf {E} \right) \end{align}$$ While this is messy to write out in general, at the node planes all the components of $\mathbf{E}$ and $\mathbf{B}$ are zero, and so $\mathbf{B}'=0$ and $\mathbf{E}'=0$.
Note that since $v$ is perpendicular to $\mathbf{E}$ and $\mathbf{B}$, the previous spatial planes where $\mathbf{E}=0$ independent of time, or $\mathbf{B}=0$ independent of time, do not maintain this nice form after transformation as the electric and magnetic parts of the field will mix in the transformation. The only truly coordinate system independent 'nodes' are the ones in which the full electromagnetic tensor is zero.
Now all that is left is to find the coordinates of these nodes in the new frame. Here I'll just work it out for the $a_n$ planes.
In the original frame nodes are at spacetime coordinates $x=a_n$,$t=\pi (m+1/2)=T_m$ while $y,z$ can have any value. Therefore in the new frame $$ ct' = \gamma(cT_m - \beta x_n)$$ Simultaneity is broken, so instead of all spatial planes appears at the same time, they will occur in order down the x axis. If there was a series of bells on the x-axis that rang whenever the electromagnetic tensor was zero, they would ring simultaneously at a set period in the rest frame, but in the moving frame there would be like travelling waves of the bells going off along the x axis.
Since these are discrete events, as the transformation parameter $v$ is increased, there would eventually be aliasing effects, making it difficult to fully interpret as a moving "wave" of nodes.
$$ x' = \gamma x_n - \gamma \beta cT_m$$
So similar to the original frame, $y',z'$ can take any value (so the nodes are still planes), with $x'$ having discrete values (but now these values depend on time). With each appearance of a node $n$, it will move along the $x$ axis. As before, with larger $v$ aliasing of the discrete events makes it suspect to 'label' nodes by $n$ to watch them move along.
Some concluding remarks on the calculation:
Since we started with a superposition of two plane waves, we could also write the result in the new frame as a superposition of two transformed plane waves. Since these waves will now have mismatched spatial and time frequencies, there will no longer be nice static planes of zero electric or magnetic field, but the nodes where $\mathbf{E}=\mathbf{B}=0$ still exist and allow us to mentally picture at least some features of the solutions.