Is the normed linear space $X$ isometrically isomorphic to $Y$, if there is a linear operator $T: X \to Y$ such that $\|T\|=1$? My question is if $\|T\|=1$ a sufficient condition for isometric isomorphism. If yes, how to prove it? If not, which additional conditions are needed? Perhaps $\|T^{-1}\|=1$ is necessary? 
 A: No, it is not sufficient. Take $\operatorname{Id}\colon(\mathbb{R}^2,\|\cdot\|_1)\longrightarrow(\mathbb{R}^2,\|\cdot\|_2)$. Then $\|\operatorname{Id}\|=1$, but $(\mathbb{R}^2,\|\cdot\|_1)$ and $(\mathbb{R}^2,\|\cdot\|_2)$ are not isometric.
A: Certainly, $\|T\|=1$ is not enough. In fact if $T$ is any nonzero operator then $S=\frac 1 {\|T\|}T $ satisfies $\|S||=1$ so such an operator exists for any two spaces (except $\{0\}$).  Suppose $\|T\|=\|T^{-1}\|=1$. Then $\|x||=||T^{-1} Tx|| \leq \|Tx||$ and $\|x||=||T T^{-1}x|| \leq \|T^{-1}x||$. Changing $x $ to $Tx$ in the second inequality we get $\|Tx\| \leq ||x||$. Hence $T$ is an isometric isomorphism. Conversely, if $T$ is an isometric isomorphism then $\|T^{-1}||=1$. So we can say that a bijective linear map $T$ of norm $1$ is an isometric isomorphism iff $\|T^{-1}\|=1$.
A: It is known that $c_0$ and $c$ equipped with the $\|\cdot\|_\infty$-norm are not isometrically isomorphic.
Hovewer, the inclusion $T : c_0 \hookrightarrow c$ has norm $1$ (it is even isometric).
