Confusion in N of N-point-FFT
Firstly, you need to know that the FFT computes the DFT (discrete Fourier transform) in an efficient manner. Hence, the output of an N-point FFT and N-point DFT are exactly the same. 

Now, especially, if N is a power-of-two, the FFT can be calculated very efficiently. Though, if N can be factorized into rel. small prime numbers (2, 3, 5), still the calculation can be done very efficiently. Strictly speaking, the Radix-2 FFT only works, if N is power-of-two. However, in common implementations in software, N can be any number and the implementation chooses the appropriate algorithms for most efficient calculation.

Hence, if N is power of two, the calculation by FFT is just a bit faster, compared to when N would be a large prime number. 

Regarding sampling frequency: If $F_s$ is your sampling frequency and $T$ is the duration of the FFT window, then $N=TF_s$. It's hard to answer more detailed what you mean by relation to sampling frequency, because I dont know exactly what you are asking.

Regarding your comment: The DFT maps a subset of the frequency range $[0,\dots,F_s]$  to the $N$ DFT-bins, which are indexed by $k=0,\ldots,N-1$. Since all frequency bins in the DFT output have the same distance to each other (i.e. they are linearly spaced), you immediately end up with $f=\frac{k}{N}F_s$, i.e. the $k$th bin corresponds to the physical frequency $f=\frac{k}{N}F_s$.