Is photon redshifted if observer starts moving away after photon is released A. A light source and an oberserver are stationary relative to each other.  The source emits a photon and the observer measures the wavelength.
B. A light source and an observer are stationary relative to each other.  The source emits a photon.  After the photon is emitted and while it is still in transit, the observer begins to move away from the source. The observer measures the wavelength while moving away from the source.
Is the photon wavelength redshifted in experiment B compared to experiment A?
If yes, how?  (Since the photon travels at speed c no matter the reference frame.)
 A: Yes, the light would be redshifted in experiment B. The measured Doppler shift depends on the velocity of the receiving instrument when the light is received relative to that of the emitter when the light was emitted.
The frequency and wavelength are both changed with respect to the values they would have in the frame of reference of the emitter, but their product, the speed of light, remains the same.
You can interpret this in terms of the classical Doppler effect modified by a time-dilation term. i.e. The relativistic Doppler effect is that
$$ \lambda_{\rm obs} = \lambda \sqrt\frac{c+v}{c-v} = \lambda\left(1 + \frac{v}{c}\right)\left(1 - \frac{v^2}{c^2}\right)^{-1/2}$$
e.g. https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Relativistic_longitudinal_Doppler_effect
A: I'm assuming you mean the direction of the photon emitted is towards the observer.
According to an approximation of Doppler Effect and Special Relativity, $λ$ = ($v/c$+1)$λ_0$, where $λ_0$ is the initial wavelength of the emitted photon and $λ$ is the resultant wavelength. $v$ is the velocity of the observer and $c$ is the speed of light.
In your case, the observer is receding from the source, so $v/c$ > 0 and so $λ$ > $λ_0$. This means redshift occurred in Experiment B.
