If $UU^*=I$ is the unitary group, what is $(UU^*)^2=1$?

Question

Let $U$ be a $n\times n$ complex matrix satisfying:

$$ UU^*UU^*=I $$

Does it follow that $UU^*=I$?

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Let me have a go at it:

$$ UU^*UU^*=(UU^*)^2=I\implies UU^*=\pm \sqrt{I} $$

Now the square root of $I$ is the set of all matrices $A^2=I$.

Does this "group" have any interesting properties, notably in physics?

Answer 1

By the given condition, both $UU^\ast$ and $I$ are positive semidefinite square roots of $I$. Since each PSD matrix has a unique PSD square root, we have $UU^\ast=I$, i.e., $U$ is simply a unitary matrix.