= Solution
For $|\alpha|^2+|\beta|^2=1$, the <density operator> and its computational-basis matrix are
$$
\hat\rho=|\psi\rangle\langle\psi|,\qquad\rho=\begin{pmatrix}|\alpha|^2&\alpha\beta^*\\\alpha^*\beta&|\beta|^2\end{pmatrix}.
$$
It is a <Hermitian matrix> and satisfies $\rho^2=\rho$, so $\operatorname{tr}\rho^2=1$. Its eigenvalues are one and zero. Thus \b[a pure qubit's density matrix is not unitary]: it is singular, and $\rho^\dagger\rho=\rho\ne I$.
Transposing gives $\rho^T=|\psi^*\rangle\langle\psi^*|$. It reflects the Bloch vector $(x,y,z)$ to $(x,-y,z)$. Choose the positive axis states $|+\rangle$, $|+i\rangle=(|0\rangle+i|1\rangle)/\sqrt2$ and $|0\rangle$; their target kets are respectively $|+\rangle$, $|-i\rangle$ and $|0\rangle$.
Nonunitarity alone does not make a quantum operation unphysical. The <qubit transpose map> fails <complete positivity>: applying it to one half of a Bell pair gives a matrix with a negative eigenvalue. A physical approximation must instead be a valid <quantum channel>.
Use the <squared quantum fidelity> specified in the question, comparing each actual output to the transposed target. For a pure input Bloch vector $\mathbf r$, the identity approximation has $F_I=1-y^2$. Conjugation by the <Pauli X gate> produces $(x,-y,-z)$, giving $F_X=1-z^2$. Measuring in the computational basis and preparing the corresponding basis state is the dephasing channel $(x,y,z)\mapsto(0,0,z)$, giving $F_Z=(1+z^2)/2$. Therefore
$$
\begin{array}{c|ccc|c}
\text{approximation}&+x&+y&+z&\text{average}\\\hline
I&1&0&1&2/3\\
X&1&1&0&2/3\\
\text{measure and prepare}&1/2&1/2&1&2/3
\end{array}
$$
The stipulated three-axis averages also agree with the uniform Bloch-sphere averages, since each squared coordinate averages to $1/3$.
The complete perfect-fidelity input sets, up to a global phase, are:
* For the identity, $y=0$: all states whose two amplitudes can be chosen real, the Bloch sphere's $xz$ great circle.
* For X, $z=0$: all equal-magnitude superpositions $(|0\rangle+e^{i\phi}|1\rangle)/\sqrt2$, its equator.
* For measurement and preparation, $z^2=1$: only $|0\rangle$ and $|1\rangle$.
Each condition follows by setting the corresponding fidelity formula to one, so these lists include all perfect cases rather than just the three test inputs.
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