Is the unitary operator unique?

[Demon117]

So there is a theorem at the beginning of section 1.5 in Sakurai that states the following:

Given two sets of base kets, both satisfying orthonormality and completeness. there exists a unitary operator U such that

|b^{(1)}> = U|a^{(1)}>,|b^{(2)}> = U|a^{(2)}>,...,|b^{(n)}> = U|a^{(n)}>

By a unitary operator we mean an operator fulfilling the conditions

U^{t}U=1

as well as

UU^{t}=1

So this is not difficult to prove. But my real question is can we prove that U is unique or is that just not the case and why?

 

[Physics Monkey]

The elements of U are U_{nm} = \langle a_n | U | a_m \rangle = \langle a_n | b_m \rangle, so if you fix the basis sets then you fix U.

 

[Demon117]

The elements of U are U_{nm} = \langle a_n | U | a_m \rangle = \langle a_n | b_m \rangle, so if you fix the basis sets then you fix U.

That is what I thought, but I wanted a second opinion. Thank you.