Is the position operator the generator of any unitary group? In other words, can it be exponentiated? Using Stone's theorem of course the answer is yes to both, but I was wondering if calculating the corresponding $U(t)$ is feasible and if it has been done, and if the result has any experimental or theoretical significance.
 A: The simple answer.
In position space, the position operator $\hat x$ is simply a multiplication of the wave function with $x$. The unitary $\hat U_x(p_0) = \mathrm e^{\mathrm i p_0 \hat x / \hbar}$ can easily be calculated: It multiplies the wave function with $\mathrm e^{\mathrm i p_0 x / \hbar}$,
$$ \langle x \mid \hat U_x(p_0) \mid \psi \rangle = \langle x \mid \mathrm e^{\mathrm i p_0 \hat x / \hbar} \mid \psi \rangle = \mathrm e^{\mathrm i p_0 x / \hbar}\, \langle x \mid \psi \rangle . $$
A bit more insight.
From how you formulated the question, I assume you know that the momentum operator generates translations in position space,
$$ \langle x \mid \hat U_p(x_0) \mid \psi \rangle = \mathrm e^{\mathrm i x_0 (-\mathrm i\hbar \partial_x) / \hbar} \langle x \mid \psi \rangle = \langle x + x_0 \mid \psi \rangle = \psi(x+x_0) . $$
Now, position and momentum operator behave very symmetrically. Let us see how $\hat U_x(p_0)$ acts in momentum space on a wave function $\langle p \mid \psi \rangle = \tilde\psi(p)$:
$$ \begin{aligned}
\langle p \mid \hat U_x(p_0) \mid \psi \rangle &= \int \mathrm dx\, \langle p \mid x \rangle \langle x \mid \hat U_x(p_0) \mid \psi \rangle \\
&= \frac{1}{\sqrt{2\pi\hbar}} \int \mathrm dx\, \mathrm e^{-\mathrm i px / \hbar}\, \mathrm e^{\mathrm i p_0 x / \hbar}\, \langle x \mid \psi \rangle \\
&= \int \mathrm dx\, \langle p - p_0 \mid x \rangle \langle x \mid \psi \rangle \\
&= \langle p-p_0 \mid \psi\rangle = \tilde\psi(p-p_0)
\end{aligned}$$
We see: $\hat x$ generates translations in momentum space.
