= Solution
For <teleportation with an ambiguous Bell-state phase label>, write $\rho=|\psi\rangle\langle\psi|$. The <Bell-basis measurement> outcomes $\phi^+$ and $\phi^-$ each occur with <probability> $1/4$. In either case a correct guess gives Bob $\rho$, while the wrong guess differs by the <Pauli Z gate> and gives $Z\rho Z$. The other two <Bell states> are identified correctly. Since Alice guesses each ambiguous label with <probability> $1/2$, the overall wrong-correction <probability> is $(1/2)(1/2)=1/4$. Averaging over the actual <quantum measurement> outcome and her guess gives the <phase-flip channel>
$$
\boxed{\rho_{\rm out}=\mathcal E(\rho)=\frac34\rho+\frac14Z\rho Z
=\begin{pmatrix}|\alpha|^2&\alpha\beta^*/2\\\alpha^*\beta/2&|\beta|^2\end{pmatrix}.}
$$
Conditional only on an ambiguous analyzer result, the output is $(\rho+Z\rho Z)/2$; that conditional channel has weight $1/2$ in the displayed unconditional result. There is no rejection of those rounds.
Use the <teleportation fidelity> for a <pure state> target, namely the <squared quantum fidelity>
$$
F_\psi=\langle\psi|\rho_{\rm out}|\psi\rangle.
$$
This compares the input with the resulting <mixed state>; averaging state vectors before forming a <density matrix> would give a different, incorrect answer. Direct substitution gives
$$
\boxed{F_\psi=\frac34+\frac14|\langle\psi|Z|\psi\rangle|^2
=\frac34+\frac14(|\alpha|^2-|\beta|^2)^2
=1-|\alpha|^2|\beta|^2.}
$$
Thus \b[the maximum is $1$], attained by $|0\rangle$ and $|1\rangle$ up to a <global phase>, because these are <eigenstates> of the <Pauli Z gate>. \b[The minimum is $3/4$], attained by every equal-weight <quantum superposition>
$$
|\psi\rangle=\frac{|0\rangle+e^{i\theta}|1\rangle}{\sqrt2},\qquad \theta\in\mathbb R,
$$
on the equator of the <Bloch sphere>. The <unsquared quantum fidelity> is $\sqrt{F_\psi}$ if that alternative convention is used. No average over different input <quantum states> is needed for the requested state-dependent answer.
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