Let $H$ be a finite-dimensional Hilbert space with identity operator $I$. It is well-known that the von Neumann entropy $S$ of a density operator $\rho$ on $H$ is maximized if and only if $\rho = I$. That is, $$S(\rho) = S(I) \Longleftrightarrow \rho = I.$$ Is this true in general for mixed states, up to unitary equivalence? In other words, if $\rho$ and $\sigma$ are two density matrices on $H$ representing mixed states such that $S(\rho) = S(\sigma)$, does it follow that $\rho = U\sigma U^\dagger$, for some unitary $U$?
Obviously, no, for a simple counting argument: The entropy is only a single number, while states for which $\rho=U\sigma U^\dagger$ must have identical spectrum, which is specified (up to trace) by $d-1$ numbers.
So if the Hilbert space has dimension $d>2$, there are counter-examples. (Just take any two diagonal states with different entries but same entropy.)
On the other hand, for $d=2$, your statement is correct, as the spectrum is specified by one number (since the trace is one).