Exercise 6.5.23

We define $T$ to be $T(x)=f(x)-f(0)$. By the proof of Theorem 6.22, we know that $T$ is an unitary operator. Also, by Theorem 6.22 we know $f$ is surjective since it’s composition of two invertible functions. Hence we may find some element $t$ such that $f(t)=2f(0)$. Now let $g(x)=x+t$. Since $T$ is linear, we have

Finally, if $f(x)=T(x+t)=U(x+v_0)$ for some unitary operator $U$ and some element $v_0$. We’ll have

Since $T$ is unitary and hence injective, we know that $t=v_0$. And thus $U$ must equal to $T$. So this composition is unique.