= Solution
Define <orthonormal> symmetric <vectors>
$$
|s_0\rangle=|\widetilde0\rangle,\qquad |s_j\rangle=\frac{|\widetilde{-j}\rangle+|\widetilde j\rangle}{\sqrt2}\quad(1\leq j\leq N).
$$
The <symmetric-arm reduction of an exchange Hamiltonian> here has two arms. To verify it directly, the link between radial distances $j$ and $j+1$ has strength $J_j$ on either side, while the two central links have strength $J_0/\sqrt2$. The calculation of part (a) therefore gives
$$
\begin{aligned}
H_e|s_0\rangle&=J_0|s_1\rangle,\\
H_e|s_j\rangle&=J_{j-1}|s_{j-1}\rangle+J_j|s_{j+1}\rangle\quad(1\leq j<N),\\
H_e|s_N\rangle&=J_{N-1}|s_{N-1}\rangle.
\end{aligned}
$$
In particular this subspace is invariant and contains the initial <vector>. Define the <linear isometry> $F|j\rangle=|s_j\rangle$. The displayed <matrix> entries prove $H_eF=FH_T$, or $F^\dagger H_eF=H_T$. Induction gives $H_e^mF=FH_T^m$ for every nonnegative integer $m$; hence the convergent exponential series proves $e^{-iH_et}F=Fe^{-iH_Tt}$. This establishes the mapping for the entire evolution, rather than just its first two powers. The supplied endpoint transfer also forces every $J_j$ to be nonzero, since a broken link would disconnect the two endpoints. Successive powers applied to $|s_0\rangle$ then generate all $|s_j\rangle$, so this is exactly the <cyclic subspace> governing the initial state.
Using the stated <perfect quantum state transfer> identity at time $\pi/2$,
$$
\boxed{e^{-iH_e\pi/2}|\widetilde0\rangle=F|N\rangle=\frac{|\widetilde{-N}\rangle+|\widetilde N\rangle}{\sqrt2}.}
$$
The two terms have the excitation at sites $0$ and $2N$, respectively, and all intermediate <qubits> are zero in both terms. The full state thus factors into a <Bell state> at the endpoints and the vacuum on the interior. Its <partial trace> is
$$
\boxed{\rho_{0,2N}=|\Psi^+\rangle\langle\Psi^+|,\qquad |\Psi^+\rangle=\frac{|10\rangle+|01\rangle}{\sqrt2}.}
$$
The coherent cross terms survive the trace because the interior <vectors> coincide. We have used the transfer phase convention supplied in the question; a common transfer phase would leave this <density operator> unchanged.
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