Suppose you have a Hamiltonian of the form
$$ H = ZXXX + YXXX + XXXX $$
where $Z,X,Y$ are the usual Pauli matrices with $ZXXX = Z \otimes X \otimes X \otimes X$ and similar for the other two terms. Since the last three qubits are being acted upon by the same operator $XXX$ can we simplify the calculation of $\langle \psi | H | \psi \rangle = \langle \psi | (Z+Y+X) \otimes XXX| \psi \rangle$ more efficiently?
Here is a little more background to this question: Usually given a Hamiltonian $H = \sum_i \alpha_i P_i$, one can determine a Clifford unitary transformation through $\mathbb{Z}_2$ symmetries (and permutation) to transform the Hamiltonian so that the transformed Hamiltonian have the last $k$ qubits are either acted by $I$ or $X$ only. Once you get to such form then you can taper off these $k$ qubits by replacing them with their eigenvalues $\pm 1$ as stated here on page 14 of this paper Tapering off qubits to simulate fermionic Hamiltonian. I was able to find the $\mathbb{Z}_2$ symmetries to transformed the original Hamiltonian (Table B1 on pg. 13) to the transfomred Hamiltonian (Table B4 on page 14) in the linked paper, but I don't fully understand how to replace these operators with their $\pm 1$ eigenvalues to get rid of these qubits.