Suppose you have a deck of cards and a table with 52 slots for the shuffled cards.
When you do the Fischer Yates, you are basically doing this:
Select a random card, put it in slot $1$.
Of the remaining $5$1, select another random card, put it in slot $2$ and continue.
People usually are confused with Fischer Yates, because implementations typically use the deck of cards (the input array) as the set of slots itself, and the act of putting a card in a slot is done by swapping cards (the elements of the input array). This is actually done to make the algorithm $\mathcal{O}(n)$ in the input array case (basically, the 'get random card', becomes $\mathcal{O}(1)$).
The probability that card $j$ goes in slot $k$ is given by (where $n=52$)
$\frac{n-1}{n}\cdot \frac{n-2}{n-1} \cdots \frac{n-k+1}{n-k+2} \cdot \frac{1}{n-k+1} = \frac{1}{n}$
(what above computes is: does not go into slots $1$ to $k-1$ and goes in to slot $k$).
Each card is equally likely as any other card to end up in a given slot.
Thus given two permutations $A$ and $B$, both are equally likely to occur.
Note: You might think that probability of getting a specific permutation is $\frac{1}{n^n}$ (or $\frac{1}{n^{n-1}}$) based on the above probability of $\frac{1}{n}$, but it is not. Can you say why?