In Shor's algorithm, how can we guarantee that each controlled-U will kickback to the same eigenvalue?

Have you seen this document? https://qiskit.org/textbook/ch-algorithms/shor.html

Note that in Shor's algorithm, we use the quantum computer as a subroutine to essentially find the period of the function

$$ f(x) = a^x mod N$$

where $a$ is a guessed value between $1$ and $N-1$. So you have to create different circuit to implement each of the guessed $a$.

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As for the QPE step, this is essentially as follow:

Let's suppose that

$$ U|psirangle = e^{2pi i phi} |psirangle$$

then

$$U^{2^j}|psi rangle = U^{2^j -1}bigg(U|psiranglebigg) = U^{2^j -1}bigg( e^{2pi i phi} |psiranglebigg) = cdots = e^{2pi i 2^j phi} |psi rangle$$

The phase-kickback turn each of the ancilla qubit (after going through Hadamard gate) from the state $dfrac{|0rangle + |1 rangle}{sqrt{2}}$ to the state $dfrac{ |0rangle + e^{2pi i 2^j phi}|1 rangle}{sqrt{2}}$ under $CU^{2^j}$ operator. To be mathematical precise,

$$CU^{2^j}: bigg( dfrac{1}{sqrt{2}} big( |0rangle + |1rangle big) bigg)|psirangle to dfrac{1}{sqrt{2}} bigg( |0rangle |psi rangle + |1rangle e^{2pi i 2^j phi} |psirangle bigg) = dfrac{1}{sqrt{2}} bigg( |0rangle + e^{2pi i 2^j phi} |1rangle bigg)|psirangle $$

Now if you apply the inverse QFT to all the ancilla qubit then you will get the binary expression of $phi$.

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