Let $\lambda_j, j=1 \ldots k$ be the distinct eigenvalues of $U$ (which must be numbers of absolute value $1$). For each $\lambda_j$ let $\mu_j$ be a square root of $\lambda_j$. These also have absolute value $1$. There is a polynomial $p(z)$ such that $p(\lambda_j) = \mu_j$ for each $j$. Let $S = p(U)$.
1) $S^2 = p(U)^2 = U$: in fact $p(z)^2 - z$ is divisible by $\prod_j (z - \lambda_j)$, which is the minimal polynomial of $U$.
2) Since $U$ is normal, the algebra generated by $U$ and $U^*$ is commutative, and in particular $S$ is normal. Since $S$ is normal and its eigenvalues, which are the $\mu_j$, have absolute value $1$, $S$ is unitary.
3) Any nonnegative integer power of a symmetric matrix is symmetric; $S$ is symmetric because it is a linear combination of the symmetric matrices $U^j$.
4) Every matrix that commutes with $U$ commutes with each $U^j$ and therefore with $S$.