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.

"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.

We all know that and it does not adress the problem. So again, please put an explicit quantifier in front of $X$ in your post.

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