PSD estimate using FFT - scaling and units I know that similar questions have already been asked – I have read the answers, yet I am still not sure if I understand the topic properly at some points. 

I want to use FFT to calculate PSD estimate. FFT gives me a set of complex numbers that I want to transform into the PSD values. I am aware of the fact that only the first $N/2 + 1$ values are useful, and that the other half are complex conjugates of the first half. 


I know that I will need to take the magnitude squared of each (useful) complex number the FFT gave me in order to find the PSD values. I also found out that I should incorporate some scaling/normalizing of the result. It is at this point that I am unsure. In some of the replies to similar questions it is said that the magnitude squared should be multiplied by $1/N$ only, while in others it is said that it should be multiplied by $1 / \left(N\cdot F_s\right)$.  I have not been able to figure out which of these two should be used when. Can anyone explain?
The sampling frequency of my signal is $128\textrm{ Hz}$. The length of the signal to be transformed is $256$ samples. In this case I found out that the distance between the resultant values of the FFT (or PSD values) should be $128/256 = 0.5\textrm{ Hz}$. Is that correct? 
The original signal is in $\mu V$. In what values will the result of the PSD be? Is it $\mu V^2$ or $\mu V^2 /\textrm{ Hz}$? 
(There might be a connection here with what I am asking in 1), am I right?)
Have a look at this paper, section 9 goes over how to scale or normalize the FFT output. For the PSD you need to divide by the effective noise bandwidth (ENBW) which is

$$
ENBW = f_s\frac{S_2}{S_1^{2}}\\
S_1 = \sum_{j=0}^{N-1}w_j\\
S_2 = \sum_{j=0}^{N-1}w_j^{2}\\
where~w_j~is~the~jth~sample~of~the~window~function
$$