If A is a 3x3 non-singular matrix such that AA'=A'A and B=A⁻¹A', then BB' equals to

Given B=A⁻¹A'
B=A⁻¹A' ⇒ AB = A(A⁻¹A') = IA' = A'
Now, ABB' = A'B' = (BA)' = (A⁻¹A'A)' = (A⁻¹A)' = I
Therefore, ABB' = I
Premultiply by A⁻¹
A⁻¹ABB' = A⁻¹I ⇒ BB' = A⁻¹
However, this is not one of the options.
Let's reconsider:
Given B = A⁻¹A'
BB' = (A⁻¹A')(A⁻¹A')' = (A⁻¹A')(A(A⁻¹)ᵀ) = A⁻¹(A'A)(A⁻¹)ᵀ
Since AA' = A'A, then A' = A⁻¹AA'
B = A⁻¹A' = A⁻¹(A⁻¹AA')
BB' = (A⁻¹A')(A⁻¹A')' = (A⁻¹A')(A(A⁻¹)ᵀ) = A⁻¹A'A(A⁻¹ᵀ) = A⁻¹(AA') (A⁻¹ᵀ)
Since AA' = A'A, then BB' = A⁻¹(A'A)(A⁻¹ᵀ) = (A⁻¹A')(A⁻¹A')' = I