How does sampling rate of $x[n]$ relate to sampling rate in frequency domain after DFT transformation?
A better way to think about it is the DFT is agnostic about the sampling rate.  What matters is the number of sample points.  If you look at the DFT definition, $F_s$ does not appear in it, but $N$ does twice.

The sampling rate can be used to figure out the duration of the sampling frame given the number of samples, or used to figure out how many sample points for a given duration.  Once the DFT is taken on the sample, the bin number corresponds to the number of cycles per frame.  The number of cycles per frame can be divided by the number of seconds per frame (the frame duration), or whatever unit of time is being used, to get the number of cycles per second (or time unit).

I think the equation the r b-j gave is better expressed as:

$$ f = k \cdot \frac{F_s}{N} $$

Where the units are:

$$ \frac{cycles}{timeunit} = \frac{cycles}{frame} \cdot \frac{\frac{samples}{timeunit}}{\frac{samples}{frame}} $$

The latter factor can be considered a conversion factor for the frequency.

When $ k = N/2 $:

$$ f = \frac{N}{2} \cdot \frac{F_s}{N} = \frac{F_s}{2} $$

Which confirms r b-j's statement about the halfway bin corresponding to the Nyquist frequency.

Hope this helps.

Ced