Vtyjkyuk

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?

No, it is not sufficient. Take $operatornameIdcolon(mathbbR^2,|cdot|_1)longrightarrow(mathbbR^2,|cdot|_2)$. Then $|operatornameId|=1$, but $(mathbbR^2,|cdot|_1)$ and $(mathbbR^2,|cdot|_2)$ are not isometric.

Certainly, $|T|=1$ is not enough. In fact if $T$ is any nonzero operator then $S=frac 1 TT $ 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 $|x||=||T T^-1x|| leq |T^-1x||$. 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 $|T^-1|=1$.

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).