I have two matrices $A_{m\times n}$ and $B_{n \times p}$ and another similar matrix $A'$ and $B'$ of same dimensions.
I am able to show that $AB = A'B'$ and $AA^T = A'A'^T$ and $BB^T = B'B'^T$.
Can I assume that $A = A'$ and $B = B'$ are true. If not, can you please help me what other product of matrices in $A$ and $B$, I should try to proof such that the correspondence can be shown.
Take $A=diag(1,2,3),B=diag(4,5,6)$ and $A'=diag(1,2,-3),B'=diag(4,5,-6)$.
EDIT 1. When $n=2$ and $A$ is triangular, it remains $3+4+4+4=15$ unknown entries in $A,A',B,B'$. The solutions of $AB=A'B',AA^T=A'A'^T,BB^T=B'B'^T$ ($12$ relations in the entries) depend on $7$ parameters. Then there are only $15-7=8$ algebraically independent relations.
EDIT 2. Correction. You want to obtain that necessarily $A=A',B=B'$, that implies that the solutions depend on at most $3+4=7$ parameters. Finally, one should not be far from the desired result. Also, look at what happens when you change $A$ with $-A$ or $B$ with $-B$.