To complement John Doe's answer: If $U,V$ are unitary operators such that $A=U M_t U^\ast=V M_t V^\ast$, then $V^\ast U$ is a unitary on $L^2(\sigma(A),E_x)$ such that
$$
V^\ast UM_t=M_t V^\ast U.
$$
Iterating this identity, we get $V^\ast UM_{t^k}=M_{t^k}V^\ast U$ for all $k\in\mathbb N$, and then by linearity and continuity also $V^\ast U M_f=M_f V^\ast U$ for all $f\in C(\sigma(A))$.
Thus, with $u=V^\ast U(1)$ we get
$$
V^\ast U f=V^\ast U M_f (1)=M_f V^\ast U(1)=uf
$$
for $f\in C(\sigma(A))$. By density, the same identity holds for all $f\in L^2(\sigma(A),E_x)$. Hence $V^\ast U=M_u$. From the fact that $V^\ast U$ is unitary, it follows easily that $u=1$ a.e.
Thus, given one unitary operator $U$ such that $A=U M_t U^\ast$, every other unitary operator $V$ that satisfies the same identity is of the form $V=U M_{u}$ with $u\colon \sigma(A)\to\mathbb T$ measurable (if you do the computations, you'll see that this $u$ is actually $\bar u$ from above).