= Solution
Use the convention that the <inner product> is linear in its first argument. The precise <polarization identity> is
$$
\boxed{\langle x,y\rangle=\frac14\left(
\|x+y\|^2-\|x-y\|^2+i\|x+iy\|^2-i\|x-iy\|^2\right).}
$$
If (1) holds, apply this formula before and after $U$ to get preservation of all <inner products>, proving (2). If (2) holds, the identity $\langle Ux,Uy\rangle=\langle x,U^*Uy\rangle$ gives $U^*U=I$. For $z=Ux$, surjectivity then gives $UU^*z=UU^*Ux=Ux=z$, proving (3).
From (3), $\|Ux\|^2=\langle x,U^*Ux\rangle=\|x\|^2$ and likewise $\|U^*x\|=\|x\|$, proving (4). Finally, if (4) holds, apply the same <polarization identity> to $U^*$ to obtain $UU^*=I$. This makes $U$ onto; the other <norm> identity supplies the remaining requirement of (1). Thus
$$
\boxed{(1)\iff(2)\iff(3)\iff(4).}
$$
These four statements are conditions in a single equivalence proof, rather than four independent source questions.
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