What if $X$ is the zero matrix?

What if $B=-A$?

Do you mean one specific X or all possible X?

Quantifiers are essential in mathematics. Does $AX=0$ for some particular $X$ or for all $X$?

If $X=Id$ (the identity matrix), the equation is true for any $A,B \in O(n)$.

If $A,B$ symmetric and definite then $$ AA^T=BB^T $$ implies $A=\pm B$ As it is shown here. Similarly for any $A,B$ skew symmetric and definite the two matrices are equal up to a sign. So if you meant that the equality holds for all the matrices $X$ then in the symmetric definite case you have equality up to a sign

@SeverinSchraven X can not be the zero matrix, moreover X is symmetric.

@pu239 Then you should specify this in your post. Also, do you mean for all $X$ of a certain type, or just some $X$? You should try to be more precise. In any case, $A=-B$ gives a counterexample for $A\neq 0$.

@SeverinSchraven I've updated the question according to all the comments

This is not a duplicate of the linked question because there is an unknown matrix $X$ in this question in the middle.

@lisyarus not for a specific $X$, however, we know it can't be zero

@pu239 It is still unclear what you are asking. What space does $X$ belong to? Also, Marco is not suggesting that it is a dublicate. He is just answering the question if $X$ is the identity matrix.

$X$ belongs to the space of all symmetric metrics with values in (0,1). But I doubt the range of these values is important.

As for the duplicate, I was referring to someone marking this question as a duplicate of another question, not to Marco.

"the question is not for a specific X" is still unclear. Please put quantifiers in your question, as requested by many comments. Are you assuming $AXA^T = BXB^T$ for every $X$ in "the space of all symmetric "metrics" (matrices) with values in (0,1)", or for some such $X$?

@AnneBauval $X$ belongs to the space of symmetric matrices with values in (0,1), i.e. is some matrix in that space. Not a specific $X$ means it is not necessarily the identity matrix, although that is one case.

I have already written "given square matrices $A, B, X$" with $X \in$ some set $\mathcal{S}$ i.e. $X$ is some matrix from that set, not $ \forall X \in \mathcal{S}$

Then the answer to your question is "no": you have been given counterexamples.

Counterexamples can be chosen as $X \gets \text{diag}(1,0,...,0)$ and $A$ and $B$ only have their first column to be the same, while other columns are different. Then it still holds that $AXA^T = B X B^T.$ But $A \neq B$ by construction.

Thank you everyone for your insights