I think you might try approaching this in the Heisenberg picture.
The time derivative of the position operator is:
$$\dfrac{d \hat x}{dt} = \dfrac{i}{\hbar}[\hat H, \hat x]$$
which is a reasonable velocity operator. The time derivative of the velocity operator is then:
$$\dfrac{d^2 \hat x}{dt^2} = \dfrac{i}{\hbar}[\hat H, \dfrac{d \hat x}{dt}]$$
For example, consider a free particle so that $\hat H = \frac{\hat P^2}{2m}$. The velocity operator would then be $\frac{\hat P}{m}$. This certainly looks reasonable as it is of the form of the classical $\vec v = \frac{\vec p}{m}$ relationship.
But, note that the velocity operator commutes with this Hamiltonian so the commutator in the definition of the acceleration operator is 0. But that is what it must be since we're assuming the Hamiltonian of a free particle which means there is no force acting on it.
Now, consider a particle in a potential so that $\hat H = \frac{\hat P^2}{2m} + \hat U$. The velocity operator, for this system, is then $\frac{\hat P}{m} + \frac{i}{\hbar}[\hat U, \hat x]$.
Assuming the potential is not a function of momentum, the commutator is zero and the velocity operator is the same as for the free particle.
The acceleration operator is then $\dfrac{i}{\hbar}[\hat U, \frac{\hat P}{m}]$.
In the position basis, this operator is just $\frac{-\nabla U(\vec x)}{m} $ which looks like the acceleration of a classical particle of mass m in a potential given by $U(\vec x)$.
If you represented the ddθ operator for example "under the parity operation" (what does that mean?) as a matrix, wouldn't it be the exact same matrix?
When we talk about an operator undergoing some unitary transformation, be it spatial inversion, rotation, time reversal, etc., we are saying
$$U^{-1}AU = \,\,?$$
Saying an operator is odd/even means
$$U^{-1}AU = \pm A$$
with $+$ for even and $-$ for odd.
When you look at the matrix elements of a transformed operator,
$$\langle \alpha | U^{-1}AU | \beta \rangle$$
you can see that these are not the same as those of the original operator,
$$\langle \alpha | A | \beta \rangle$$
So the matrix is not the same.
It's just that the input wavefunction that is being input as argument to the ddθ operator has had all its + and − signs flipped
That's a valid way of looking at it as well. Instead of viewing $\langle \alpha | U^{-1}AU | \beta \rangle$ as some new operator $A' = U^{-1}AU$ acting on the old states, you can see it as $A$ acting on the transformed states $\langle \tilde \alpha | = \langle \alpha | U^{-1}$ and $| \tilde \beta \rangle = U | \beta \rangle$. I'll stress that the matrix is still not the same, provided you're using the same basis in either case.
To illustrate this point, let's consider the expectation value of an operator $A$ - which commutes with the position operator - after a parity transformation. We have
$$\langle A' \rangle = \langle \Psi | \Pi^{-1}A\Pi | \Psi \rangle = \int d\mathbf{x'}d\mathbf{x''} \langle \Psi | \Pi^{-1} | \mathbf{x'} \rangle \langle \mathbf{x'} | A | \mathbf{x''} \rangle \langle \mathbf{x''} | \Pi | \Psi\rangle = \int d\mathbf{x'} \Psi^*(-\mathbf{x'})A\Psi(-\mathbf{x'})$$
where in the last step I've made use of the orthogonality of position states and the Hermiticity of the parity operator.
So either way is valid (although the relation $U^{-1}AU = \pm A$ is independent of basis). In the same way that we can speak of time-evolved operators with time-independent kets in the Schrödinger picture and time-evolved kets with time-independent operators in the Heisenberg picture, we can choose to transform the "inputs" or the operator.
Best Answer
No we cannot, since the only requirement$$\mathscr{P}^{-1}\hat{\textbf{x}}\mathscr{P}=-\hat{\textbf{x}}$$ does not fix the parity operator uniquely even in the simplest case. Further information with the form of added requirements is necessary to fix the parity operator.
The definition of parity operator actually depends on the system you are considering. Let us consider the simplest spin-zero particle in QM.
Its Hilbert space is (isomorphic to) $L^2(\mathbb R^3)$.
Parity is supposed to be a symmetry, so in view of Wigner's theorem, it is an operator $H: L^2(\mathbb R^3) \to L^2(\mathbb R^3)$ which may be either unitary or antiunitary.
Here the parity operator is fixed by a pair of natural requirements, the former is just that in the initial question, the latter added requirement concerns the momentum operators. $$UX_kU^{-1} =-X_k\quad, k=1,2,3 \tag{1}$$ and $$UP_kU^{-1} =-P_k\quad, k=1,2,3 \tag{2}$$ Notice that (2) is independent from (1), we could define operators satisfying (1) but not (2).
First of all, these requirements decide the unitary/antiunitary character. Indeed, from CCR, $$[X_k,P_h] = i\delta_{hk}I\tag{3}$$ we have $$U[X_k,P_h] U^{-1} = \delta_{kh} Ui IU^{-1} = \pm i \delta_{kh}I$$
that is $$[UX_kU^{-1}, UP_hU^{-1}]=\pm i \delta_{kh}I\:,$$ so that, from (1) and (2), $$[X_k,P_h] = \pm i \delta_{kh}I$$ Comparing with (3), this identity rules out the minus sign corresponding to an antiunitary operator. $U$ must be unitary.
Let us prove that (1)-(2) fix $U$ up to a phase. It is not possible to define $U$ more precisely because this arbitrary phase is just the degree of freadom permitted by Wigner's theorem in defining symmetries in terms of unitary or anti unitary operators.
Suppose that, for another unitary operator $V$, we also have $$VX_kV^{-1} =-X_k\quad, k=1,2,3 \tag{1'}$$ and $$VP_kV^{-1} =-P_k\quad, k=1,2,3 \tag{2'}\:.$$ As a consequence of (1) and (2), $$U^{-1}VX_kV^{-1}U =X_k\quad, k=1,2,3 $$ and $$U^{-1}VP_kV^{-1}U = P_k\quad, k=1,2,3 \:.$$ In other words, $L:= U^{-1}V$ satisfies $$L X_k = X_kL\:, \quad L P_k = P_kL\quad, k=1,2,3 \:.$$ Since the system of operators $X_k$ and $P_k$ is irreducible in $L^2(\mathbb R^3)$, Schur's lemma implies that $$L= e^{i\gamma}I$$ for some fixed real $\gamma$. Namely, $$V = e^{i\gamma}U\:.$$
To conclude, it is enough to find an unitary operator satisfying both (1) and (2). Per direct inspection one sees that $$(U\psi)(x) = \psi(-x) \tag{4}$$ does the job. All remaining possibilities are included in the arbitrary phase $e^{i\gamma}$. Choice (4) has a nice further property shared with only the other possibility $$(U\psi)(x) = -\psi(-x) \tag{4'}$$ In both situations (and only for these choices of the phase), $$UU=I\:.$$ Since we already know that $U^{-1}=U^\dagger$, we conclude that $$U= U^{-1}= U^\dagger\:.$$ In other words the said choices of the phase make $U$ an observable, which plays an important role in particle physics (it can be conserved or not depending on the Hamiltonian).
Extending the notion of particle by including the spin, the Hilbert space enlarges to $L^2(\mathbb R^3)\otimes {\mathbb C}^{2s+1}$, where the second factor includes an irreducible representation of $SU(2)$ generated by the three spin operators $S_k$, $k=1,2,3$. An analysis similar to the previous one can be developed by adding to (1) and (2) the further requirement $$U S_kU^{-1} = S_k\:, \quad k =1,2,3,$$ together with spin commutation relations $$[S_k,S_h]= i \sum_{p=1}^3\epsilon_{khp} S_p\:,$$ but I stop here.