Isometry of the adjoint operator

Question

Suppose that $T$ acts between Banach spaces $T: X \to Y$ and $T$ is surjective. One can define the adjoint operator $T^*:Y^* \to X^*$ by the formula $T^*(\varphi):=\varphi \circ T$. Is it true that if $T^*$ is an isometry then $T$ is also? If so, why is it true?

Answer 1

It is true, at least, that (given that $T$ is surjective) if $T$ is an isometry, then $T^*$ is as well. Surjectivity here is particularly important. Note that $$ \|T^*\phi\| = \sup_{x \in X \setminus \{0\}} \frac{|\phi(T(x))|}{|x|} = \sup_{x \in X \setminus \{0\}} \frac{|\phi(T(x))|}{|T(x)|} = \sup_{y \in Y \setminus \{0\}} \frac{|\phi(y)|}{|y|} = \|\phi\| $$ For the non-surjective case, take the right-shift operator as a counterexample.