Let \( H_N\vert E\rangle = E\vert E\rangle \). Thus
$$ \begin{align*} H_N e^{T} e^{-T} \vert E\rangle &= E\vert E\rangle , \\ \left(e^{-T} H_N e^T\right) e^{-T} \vert E\rangle &= Ee^{-T} \vert E\rangle , \\ \overline{H_N} e^{-T} \vert E\rangle &= E e^{-T}\vert E\rangle . \end{align*} $$Thus, if \( \vert E\rangle \) is an eigenstate of \( H_N \) with eigenvalue \( E \), then \( e^{-T}\vert E\rangle \) is eigenstate of \( \overline{H_N} \) with the same eigenvalue.