Suppose that $T$ is a bounded linear operator on a complex Banach space X and that we know that $\sigma(T) = \{1\}$ and $\|T\| = 1$ (i.e. the spectrum of the contraction $T$ consists only of a single point, 1). Does it follow that $T$ is the identity operator?
This is true in finite dimensions. In finite dimensions, the operator $N := T - \mathbb{1}$ is nilpotent. If $N\neq 0$, then there exists a strictly positive interger $D$, such that $N^D \neq 0$ and $N^{D+1} = 0$. For $K \geq D$, we have $$1 = \|T\| = \|\mathbb{1} +N\| = \|(\mathbb{1} + N)^K\| = \|\mathbb{1} + \sum_{i = 1}^K {K\choose i} N^i\| = \|\mathbb{1} + \sum_{i = 1}^D {K\choose i} N^i\|.$$ Choose a vector $x \in X$ such that $N^Dx \neq 0$, then the vectors $x, Nx, N^2x, \dots N^Dx$ are linearly independent. The coordinate function of $Nx$ is ${K\choose 1} = K$, which is unbounded as $K \rightarrow \infty$. This contradicts that $\|T\| = 1$.
In infinite dimensions, the difficulty is that $N$ is not nilpotent but merely quasinilpotent and that the coordinate functions may not be continuous. At the moment I can neither prove this nor construct a counter example.
The answer is no, it does not have to be the identity even in Hilbert space. One counterexample can be constructed as follows.
Let $H^2$ be the Hardy space in the unit disc and $S$ be the unilateral shift, i.e. $Sf=zf$, $f\in H^2$. Then the classical functional model for a completely nonunitary contraction (aka class $C_0$) is $T=P_KS|K$ where $K=H^2\ominus\Theta H^2$, $\Theta$ is an inner function and $P_K$ is the orthogonal projection on $K$. It is the known fact that $\sigma(T)$ is the spectrum of $\Theta$ (zeros and limiting zeros). If we take the singular inner function $$ \Theta_s(z)=\exp\left(\frac{z+1}{z-1}\right) $$ then the only point of spectrum becomes $z=1$. The norm inequality for the contraction $\|T\|\le 1$ becomes equality as the spectral radius is $1$. It is also known in the literature as a unicellular operator or a Jordan block.
More about the functional model can be found in
In particular, in [3,Ch.IV.3] one can find another example. Let $$ Vf(x)=\int_0^x f(t)\,dt,\quad f\in L^2[0,1], $$ (it is a Volterra operator, i.e. $\sigma(V)=\{0\}$) and $T=(I-V)(I+V)^{-1}$. Then $T$ is unitary equivalent to $P_K\Theta_s|K$ above. Alternatively, this $T$ is the cogenerator of a strongly continuous semigroup of contractions ${\cal T}(t)$ that vanishes after $t=1$ (see [3]).