Rigid bricks imply brick-finiteness?
Determine whether, for any finite-dimensional algebra A over an algebraically closed field, the condition that every brick X satisfies Ext^1_A(X,X)=0 implies that A is brick-finite (i.e., has only finitely many bricks up to isomorphism).
References
For instance, it is known that the correctness of the 2nd bBT conjecture gives an affirmative answer to the following question, which (to our knowledge) is still an open problem. If every $X\in \brick(A)$ is rigid, then is algebra $A$ necessarily brick-finite?
                — On the bricks (Schur representations) of finite dimensional algebras
                
                (2508.11789 - Mousavand et al., 15 Aug 2025) in Question [Rigid bricks], Section 4.2