The Doppler shift and the apparent speed of galactic rotation with distance A Doppler redshift would also give the illusion that galaxies were rotating more slowly then they are with the degree of illusory slowing in proportion to the degree of redshift. Do more distant galaxies appear to be rotating more slowly than closer in galaxies on average, and would this not constitute additional evidence for expansion, or if not would it not constitute evidence against expansion?
 A: Short answer: There is no test here. The effect you are talking about is already incorporated into the formula for relativistic Doppler shift and hence in calculations of the rotation speeds of distant galaxies. The inferred rotation speed depends on the ratio of a frequency difference to the observed frequency and this ratio does not change with redshift, no matter whether that overall redshift is caused by the expansion of the universe or a Doppler motion (or indeed a gravitational redshift).
In special relativity, the effects of time-dilation are already incorporated into the formula for the relativistic Doppler shift. Thus in the co-moving rest frame of the galaxy, the frequency difference between light from one side of the galaxy's rotation axis and the other is given by (ignoring inclination effects and assuming axial symmetry)
$$ \Delta f = \left(\frac{1+\beta}{1-\beta}\right)^{1/2}f_0 -\left(\frac{1+\beta}{1-\beta}\right)^{1/2}f_0 = \gamma(1 + \beta)f_0 - \gamma(1-\beta)f_0 =  2\gamma \beta f_0,$$
where $\beta = |v|/c$, $\gamma = (1 - \beta^2)^{-1/2}$ and $|v|$ is the fixed tangential rotation speed in the co-moving frame. If you measured $\Delta f$ for a line with intrinsic frequency $f_0$, then you would deduce $v$ as
$$ v = \frac{\Delta f/2f_0}{\left(1 + (\Delta f/2f_0)^2\right)^{1/2}} c$$
If the galaxy is then observed from some other frame of reference receding at velocity $v_g$ with respect to the galaxy's rest frame, the relativistic velocity addition formula can be used to show that
$$ \Delta f_g = \gamma(1 + \beta)\gamma_g (1-\beta_g)f_0 - \gamma(1-\beta)\gamma_g (1-\beta_g)f_0 = 2\gamma \beta \gamma_g(1-\beta_g)f_0, $$
where $\beta_g = |v_g|/c$.
Now the definition of redshift $z$ (whether cosmological or due to a Doppler effect) is
$$1+z = \left( \frac{1+\beta_g}{1-\beta_g}\right)^{1/2} = \gamma_g (1 + \beta_g) , $$
so
$$ (1+z)^{-1} = \gamma_g (1 - \beta_g)$$
and
$$\Delta f_g = \frac{2\gamma \beta f_0}{1+z} = \frac{\Delta f}{1+z} = 2\gamma \beta f_g,$$
where the observed central frequency of a line becomes $f_g = f_0/1+z$.
What is most important for answering your question is that the inferred rotation velocity $v$ depends only on the ratio of $\Delta f_g$ to the observed line centre $f_g$ - and this ratio is invariant, $\Delta f/f_0 = \Delta f_g/f_g = 2 \gamma \beta$ in both cases.
Now consider the same galaxy with an apparent doppler effect caused by the expansion of the universe. The frequency of light from one side of the galaxy is $\gamma (1+ \beta) f_0/(1+z)$, whilst the frequency from the other side of the rotation axis is $\gamma (1-\beta) f_0/(1+z)$ and the difference between these is $2\gamma \beta f_0/(1+z)= 2\gamma \beta f_g$ again.
Thus, there is no distinguishing test here. If galaxies have uniform rotation rates with redshift (i.e. $\beta$ and $\Delta f$ are constant) then we expect $\Delta f_g$ to decrease with increasing redshift, whether that redshift is due to the expansion of the universe or that the galaxy is simply moving away from us quickly. However the ratio of $\Delta f_g/f_g$ stays the same.
