For each Hausdorff compact $X$ and continuous map $\phi:X\to X$ denote by $\mathcal{C}(\phi)$ the operator $\mathcal{C}(\phi):C(X)\to C(X):f\to f\circ\phi$. It is straightforward to check that
$$
\mathcal{C}(\phi)\mathcal{C}(\psi)=\mathcal{C}(\psi\circ\phi)\qquad\qquad
\mathcal{C}(1_X)=1_{C(X)}
$$
Now assume that $\phi$ is a homeomorphism, id est invertible with continuous inverse. Then
as an easy consequence
1) $\mathcal{C}(\phi)$ is invertible and $\mathcal{C}(\phi)^{-1}=\mathcal{C}(\phi^{-1})$
2) for all $f\in C(X)$ we have $
\Vert \mathcal{C}(\phi)(f)\Vert_\infty
=\max_{y\in \phi(X)}|f(y)|
=\max_{y\in X}|f(y)|
=\Vert f\Vert_\infty
$
i.e. $\mathcal{C}(\phi)$ is an isometry so $\Vert \mathcal{C}(\phi)\Vert=1$
3) spectrum of every isometry on a $C^*$-algebra (and in parrticlar on $C(X)$) is contained in the unit circle. See theorem $3.1$ here.
In your particular case you need to put $X=[0,1]$ and consider homeomorphism $\phi(t)=t^2$. Then $A=\mathcal{C}(\phi)$, and it is remains to apply results above.