= Solution
An antilinear involution need not preserve norms, so the required physical assumption is that momentum reversal maps a complete normalized basis to a normalized basis. For a spinless sector take $\hat T|\boldsymbol p\rangle=\eta(\boldsymbol p)|-\boldsymbol p\rangle$ with $|\eta|=1$ and relativistic normalization $\langle\boldsymbol p|\boldsymbol q\rangle=(2\pi)^3 2E_p\delta^3(\boldsymbol p-\boldsymbol q)$. Any extra degeneracy labels must likewise be mapped by a norm-preserving permutation or unitary matrix. The condition $\hat T^2=1$ requires $\eta(-\boldsymbol p)\eta(\boldsymbol p)^*=1$ in the scalar basis.
For arbitrary normalizable wave packets $|\phi\rangle=\int d\Pi_p\,f(\boldsymbol p)|\boldsymbol p\rangle$ and $|\psi\rangle=\int d\Pi_p\,h(\boldsymbol p)|\boldsymbol p\rangle$, where $d\Pi_p=d^3p/[(2\pi)^3 2E_p]$, antilinearity conjugates their coefficients. Using the normalized reversed basis gives
$$
\langle\hat T\phi|\hat T\psi\rangle=\int d\Pi_p\,f(\boldsymbol p)h(\boldsymbol p)^*|\eta(\boldsymbol p)|^2
=\left[\int d\Pi_p\,f(\boldsymbol p)^*h(\boldsymbol p)\right]^*.
$$
Completeness proves \b[$\langle\hat T\phi|\hat T\psi\rangle=\langle\phi|\psi\rangle^*$ for all states], so this <quantum time-reversal operator> is <antiunitary>. The same proof with sums over extra labels uses the unitarity of their reversal matrix.
For a Hermitian interaction-picture Hamiltonian, the condition of <time-reversal symmetry> is
$$
\boxed{\hat TV(t)\hat T^{-1}=V(-t),\qquad V(t)^\dagger=V(t).}
$$
Expand the <scattering matrix> as a <Dyson series>, using ordered integration rather than commuting operators:
$$
S=1+\sum_{n\geq1}(-i)^n\int_{t_1>\cdots>t_n}dt_1\cdots dt_n\,V(t_1)\cdots V(t_n).
$$
Conjugation by $\hat T$ conjugates $(-i)^n$ to $i^n$ but preserves the product order. Substitute $s_j=-t_j$; the integration domain becomes $s_1<\cdots<s_n$ and
$$
\hat TS\hat T^{-1}=1+\sum_{n\geq1}i^n\int_{s_1<\cdots<s_n}ds_1\cdots ds_n\,V(s_1)\cdots V(s_n).
$$
On the other hand, taking the adjoint of the original series conjugates its coefficient and reverses the operator order. Relabel $s_1=t_n,\ldots,s_n=t_1$ in each adjoint integral; it is exactly the same series. Thus
$$
\boxed{\hat TS\hat T^{-1}=S^\dagger.}
$$
Equivalently, <time ordering> turns into anti-<time ordering> after reversing the time arguments. One can first use a finite symmetric time interval and the same time-symmetric switching prescription before taking the scattering limit; the argument never assumes $[V(t),V(s)]=0$.
<Antiunitarity> relates amplitudes by
$$
\langle\hat Tf|\hat TS\hat T^{-1}|\hat Ti\rangle=\langle f|S|i\rangle^*.
$$
Using $\hat TS\hat T^{-1}=S^\dagger$ and taking a complex conjugate gives the <time-reversal reciprocity of the scattering matrix>
$$
\boxed{\langle f|S|i\rangle=\langle\hat Ti|S|\hat Tf\rangle.}
$$
The reverse process starts with the reversed final state and ends with the reversed initial state. Intrinsic state phases can appear when these reversed states are rewritten in a separately chosen basis, but the corresponding squared amplitudes agree. Equal amplitudes do not mean identical numerical cross sections when the reversed process has different flux or phase-space factors.
For a <spin>-zero <complex scalar field>, choose the transformation convention $\hat T\phi(t,\boldsymbol x)\hat T^{-1}=\eta_T\phi(-t,\boldsymbol x)$. Its <mode expansion of a free field> is
$$
\phi(t,\boldsymbol x)=\int d\Pi_p\left[a(\boldsymbol p)e^{-iE_pt+i\boldsymbol p\cdot\boldsymbol x}
+b^\dagger(\boldsymbol p)e^{iE_pt-i\boldsymbol p\cdot\boldsymbol x}\right].
$$
Apply $\hat T$ to the expansion. Its <antiunitarity> conjugates both numerical plane waves, giving coefficients $\hat Ta\hat T^{-1}$ multiplying $e^{iE_pt-i\boldsymbol p\cdot\boldsymbol x}$ and $\hat Tb^\dagger\hat T^{-1}$ multiplying $e^{-iE_pt+i\boldsymbol p\cdot\boldsymbol x}$. In $\eta_T\phi(-t,\boldsymbol x)$, replace $\boldsymbol p$ by $-\boldsymbol p$ to obtain precisely these same plane waves. Comparing independent modes proves <time reversal of a complex scalar field>:
$$
\boxed{\hat Ta(p)\hat T^{-1}=\eta_Ta(p_P),\qquad
\hat Tb^\dagger(p)\hat T^{-1}=\eta_Tb^\dagger(p_P),\quad p_P=(E_p,-\boldsymbol p).}
$$
Taking adjoints gives $\hat Tb(p)\hat T^{-1}=\eta_T^*b(p_P)$. Positive energy is preserved, and an <annihilation operator> remains an <annihilation operator>. Time reversal does not exchange particles with <antiparticles>.
The coefficient $\eta_T$ is an <intrinsic time-reversal phase>. Applying $\hat T$ twice gives $\hat T^2a\hat T^{-2}=\eta_T^*\eta_Ta$, so $|\eta_T|=1$; it need not be restricted to $\pm1$. Write $\eta_T=e^{i\vartheta}$ and redefine $a'=e^{i\vartheta/2}a$. Because the phase is conjugated by $\hat T$,
$$
\hat Ta'\hat T^{-1}=e^{-i\vartheta/2}\eta_Ta(p_P)=a'(p_P).
$$
Thus \b[the scalar intrinsic phase can be absorbed into the annihilation-operator convention]. Redefining $b'=e^{-i\vartheta/2}b$ simultaneously implements the corresponding whole-field rephasing $\phi'=e^{i\vartheta/2}\phi$ and removes the phase from both mode transformations.
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