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?