Let $E$ be a normed space. Let $T:E\to E$ be an injective continuous linear map, such that $T^{*}$ is an isometry. Does it follows that $T$ itself is an isometry (in fact it is then an isometric isomorphism)?
This is true if $E$ is reflexive. Indeed, if $T$ is injective, $T^{*}$ has a dense range. An isometry with a dense range must be an isometric isomorphism, and so $T=T^{**}$ is also an isometric isomorphism. However, if $E$ is not reflexive we cannot conclude that $T^{*}$ has a dense image, only weak* dense, and so I expect that there is a counterexample.
Best Answer
Edit: This answer was for an older version of the question where $T$ was not assumed injective.
Use $E = c_0$ with sup norm and $T$ as the left-shift map, so $E^* = \ell^1$. Then, if $y=(y_1, y_2, \dots) \in \ell^1$, we have $T^*(y) = (0,y_1,y_2, \dotsc)$ which is clearly an isometry on $\ell^1$, but $T$ is not an isometry.