Linear isometry and operator norm $=1$

For example, on a finite dimensional space, $\|T\|=1$ if and only if for all $x$, $\|Tx\|\leq \|x\|$, and there exists $x_0\neq 0$ such that $\|Tx_0\|=\|x_0\|$. On the other hand, for $T$ to be an isometry requires for all $x$, $\|Tx\|=\|x\|$. You can give many examples of the former that do not satisfy the latter if the dimension is greater than $1$, as paul garrett commented, even diagonal matrices as Qiaochu Yuan commented. The former condition even allows $T$ to have nontrivial kernel as Martin Sleziak commented.

As David Mitra commented, if $T\neq 0$ then $\dfrac{T}{\|T\|}$ has norm $1$, and note that if there exist nonzero vectors $x$ and $y$ such $\dfrac{\|Tx\|}{\|x\|}\neq \dfrac{\|Ty\|}{\|y\|}$, then $T$ is not a multiple of an isometry. For example, if $T$ is nonzero and noninjective, take $x\in\ker T$ and $y\not\in\ker T$.