The well-known Banach-Mazur theorem says that $C([0, 1])$ is a universal separable Banach space, in the sense that if $X$ is any separable Banach space then there is a map $f : X \to C([0, 1])$ which is both linear and isometric. Note that $C([0, 1])$ also has the structure of a Banach algebra.
My question is this: Is $C([0, 1])$ universal for separable commutative Banach algebras? Of course a separable Banach algebra is a separable Banach space, so there is a linear isometry into $C([0, 1])$, but I'm asking if that map can also be taken to preserve the multiplication operation.
If $C([0, 1])$ is not universal for separable commutative Banach algebras, does there exist such a universal object? I'm interested mostly in ZFC results, but would also not mind hearing consistent answers (especially if they are consistent with $\neg CH$).