## Do we have $BTT^*B^*=B^*TT^*B$ when $T$ is closed and $B$ is normal?

I would like to know if we have the following equality $$B(TT^*)B^*=B^*(TT^*)B$$ when $$B\in B(H)$$ is normal and $$T$$ is a densely defined closed operator?

I don’t know if this can help but $$TT^*$$ is self-adjoint (because $$T$$ is closed).

If this is true, can someone give me a hint to prove it? Otherwise, can someone help me finding a counterexample?

Thank’s.