I gather you want the "seat of the pants" beastie:
$$\bbox[yellow]{\hat P= \exp \left ( \frac{-\pi}{2\hbar}(\hat {x}\hat p+\hat {p} \hat {x}) \right )}.$$
This is clearly hermitean, $\hat P ^\dagger = \hat P$, but also unitary, $\hat P ^{-1}=\hat P ^\dagger =\hat P$ : compose the exponentials in $\hat P ^2= \exp \left ( \frac{-\pi}{\hbar}(\hat {x}\hat p + \hat p \hat {x} ) \right )=1\!\! 1 $.
Given $[\hat x \hat p, \hat x]=-i\hbar \hat x $, it is evident that
$$
e^{-\pi \hat x \hat p /\hbar} f(\hat x) ~e^{\pi \hat x \hat p /\hbar} =
f(e^{-\pi \hat x \hat p /\hbar} ~\hat x ~e^{\pi \hat x \hat p /\hbar} )= f(e^{-[\pi \hat x \hat p /\hbar ,~\bullet}~~\hat x)=f(e^{i\pi} \hat x) =f(-\hat x).$$
$[A,\bullet \equiv \operatorname{ad}_A$ so that $e^A B e^{-A}= e^{[A,~\bullet} ~ B\equiv B+[A,B]+[A,[A,B]]/2!+...$, the Hadamard identiy.
It should be apparent that the same works with the full hermitean exponent, and for arbitrary functions of $\hat p$ as well. The $\hat P ^2$ expression plugged in preserves all operators.
Recall that $\hat {p}|z\rangle= i\hbar \partial_z |z\rangle$, so the resulting pseudo-dilatation operator merely flips the sign of the space argument of the ket,
$$\hat P |z\rangle=\exp (-i\pi z\partial_z +i\pi /2)|z\rangle=i|-z\rangle,$$ the i phase being an immaterial consequence of the conventions adopted here.
In operator language, in the x-representation, the pseudodilatation presents as a trivial change of variable $x=\log z$ application of Lagrange's shift operator, $\exp (a\partial_x) f(x)=f(a+x)$, namely
$$
\hat P f(z) \hat P^\dagger = e^{i\pi ~ z\partial_z } f(z) =
e^{i\pi \frac{\partial}{\partial \log z}} f(e^{\log z })=f(e^{i\pi +\log z })= f(-z).
$$
- PS: There is an alternative expression, the Weyl-ordering of the above in terms of a double parametric integral, but presumably you'd have no use for it here. An alternative, circular, formal representation of it is also $\hat P= \int dx ~|-x\rangle \langle x|$ .