Background: Let $X$ be a Banach space. We know a linear map $h$ is a surjective isometry of $X$ if and only if its adjoint $h^*$ is a surjective isometry of $X^*$.

In general, a linear map $g:X^* \to X^*$ need not have a pre-adjoint. But what if $g$ is a surjective isometry? Must there exist $f:X \to X$ such that $g=f^*$? If we identify $X$ with its embedding in $X^{**}$, this is equivalent to $g^*(X) \subset X$.

Sorry if this question is trivial; it seems like this should be well-known, but I haven't been able to find a reference or an easy counter-example.