If $$A$$ is a $$3 \times 3$$ non-singular matrix such that $$AA' = A'A$$ and $$B = A^{-1}A'$$, then $$BB'$$ equals, where $$X'$$ denotes the transpose of the matrix $$X$$.

We start with the information that $$A$$ is a non-singular (invertible) $$3\times3$$ matrix which satisfies the commutation relation $$AA' = A'A$$, where $$A'$$ denotes the transpose of $$A$$. Because $$A$$ is non-singular, all of its algebraic manipulations involving inverses are valid.

Next, the matrix $$B$$ is defined as $$B = A^{-1}A'$$. Our aim is to find $$BB'$$, the product of $$B$$ with its own transpose.

First we need an explicit expression for $$B'$$. We recall two standard facts of matrix algebra and state them clearly:

1. Transpose of a product: $$(MN)' = N'M'$$ for any two conformable matrices $$M$$ and $$N$$.

2. Transpose of an inverse: $$(M^{-1})' = (M')^{-1}$$ for any invertible matrix $$M$$.

Applying fact 1 to $$B = A^{-1}A'$$, we have

$$ B' \;=\; (A^{-1}A')' \;=\; (A')' (A^{-1})'. $$

But $$(A')' = A$$ because taking the transpose twice returns the original matrix, and by fact 2 we have $$(A^{-1})' = (A')^{-1}.$$ Substituting these two results gives

$$ B' \;=\; A\, (A')^{-1}. $$

Now we multiply $$B$$ by this $$B'$$ to find $$BB'$$:

$$ BB' \;=\; (A^{-1}A') \, \bigl(A\,(A')^{-1}\bigr). $$

Because matrix multiplication is associative, we can regroup without changing the order of individual factors:

$$ BB' \;=\; A^{-1}\, \bigl(A'A\bigr)\,(A')^{-1}. $$

At this point, the given commutation relation $$AA' = A'A$$ becomes crucial. It tells us that $$A'A = AA'$$, so in the middle of the product we may interchange $$A'$$ and $$A$$:

$$ A'A \;=\; AA'. $$

Substituting $$AA'$$ for $$A'A$$ in the expression for $$BB'$$ yields

$$ BB' \;=\; A^{-1}\,\bigl(AA'\bigr)\,(A')^{-1}. $$

Again using associativity, we first group $$A^{-1}A$$, which collapses to the identity matrix $$I$$ because $$A^{-1}$$ is the inverse of $$A$$:

$$ A^{-1}A \;=\; I. $$

Hence we obtain

$$ BB' \;=\; I\,A'\,(A')^{-1}. $$

The product of a matrix with its inverse is also the identity, so $$A'(A')^{-1} = I$$. Therefore

$$ BB' \;=\; I\,I \;=\; I. $$

We have reached the result that the product $$BB'$$ equals the $$3\times3$$ identity matrix $$I$$.

Hence, the correct answer is Option D.