Equation of motion for time evolution operator in the interaction picture Consider a system with Hamiltonian $\hat{H}=\hat{H_0}+\hat{V}$. We define the interaction picture kets $|\psi(t)\rangle _I$ by
$$\tag{1} |\psi(t)\rangle _I=\exp\left(\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)|\psi(t)\rangle$$
and interaction picture operators $\hat{O}_I$ by
$$\tag{2} \hat{O}_I= \exp\left(\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)\hat{O}\exp\left(-\frac{i}{\hbar}\hat{H}_0(t-t_0)\right).$$
From the ordinary Schrödinger equation, we obtain the equation of motion for the state:
$$\tag{3} i\hbar\frac{d}{dt}|\psi(t)\rangle _I = \hat{V}_I|\psi(t)\rangle _I.$$
As is done e.g. in Sakurai's book, we also define the time-evolution operator $\hat{U}(t,t_0)_I$ by
$$\tag{4} |\psi(t)\rangle _I = \hat{U}(t,t_0)_I |\psi(t_0)\rangle _I.$$
Differentiating (4) w.r.t $t$ yields
$$\tag{5} \begin{align}
i\hbar\frac{d}{dt}|\psi(t)\rangle _I &= i\hbar\frac{d}{dt}\left(\hat{U}(t,t_0)_I\right) |\psi(t_0)\rangle \\
&=i\hbar\frac{d}{dt}\left(\hat{U}(t,t_0)_I\right)\hat{U}(t,t_0)_I^{-1} |\psi(t)\rangle.
\end{align}$$
Comparison of (3) and (5) shows that we must have the following equation of motion for $\hat{U}(t,t_0)_I$:
$$\tag{6} i\hbar\frac{d}{dt}\hat{U}(t,t_0)_I = \hat{V}_I\hat{U}(t,t_0)_I.$$
However, if we take $\hat{O}=\hat{U}(t,t_0)$ in equation (2), then equation (2) would yield
$$\tag{7} \begin{align}
&i\hbar\frac{d}{dt}\hat{U}(t,t_0)_I\\
&= \exp\left(\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)(-\hat{H}_0\hat{U}(t,t_0)+\hat{U}(t,t_0)\hat{H}_0)\exp\left(-\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)\phantom{abc}\\&+\exp\left(\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)\hat{H}\hat{U}(t,t_0)\exp\left(-\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)\\
&= \hat{V}_I\hat{U}(t,t_0)_I+\hat{U}(t,t_0)_I\hat{H}_0.
\end{align}$$
Evidently, equations (6) and (7) do not agree. Am I misunderstanding something here?
 A: The relation between the time evolution operator $U_\mathrm I$ in the interaction picture and the time evolution operator $U_\mathrm S$ in the Schrödinger picture is not given by equation $(2)$ in the OP. In other words, the operator obtained from transforming $U_\mathrm S$ into the interaction picture is not equal to the time evolution operator in the interaction picture.
In fact, the correct relation reads
$$U_\mathrm I (t,t^\prime) = e^{iH_0t}\, U_\mathrm S (t,t^\prime) \,e^{-iH_0t^\prime } \tag{1} \quad , $$
which follows from
$$|\Psi(t)\rangle_\mathrm I = e^{iH_0t}\,U_\mathrm S(t,t^\prime) \, |\Psi(t^\prime)\rangle_\mathrm S = e^{iH_0t}\,U_\mathrm S(t,t^\prime)\, e^{-iH_0t^\prime} |\Psi(t^\prime)\rangle_\mathrm I \overset{!}{=} U_\mathrm I (t,t^\prime)\, |\Psi(t^\prime)\rangle_\mathrm I\quad . $$

By using the correct relation between these evolution operators, we obtain the correct differential equation for $U_\mathrm I$. Indeed, differentiating equation $(1)$ with respect to $t$ shows that
\begin{align}
i \frac{\mathrm d U_\mathrm I (t,t^\prime)}{\mathrm d t} &= - e^{iH_0t}\,H_0\, U_\mathrm S(t,t^\prime) \, e^{-iH_0t^\prime} + e^{iH_0t} \,\overbrace{i\frac{\mathrm d U_\mathrm S (t,t^\prime)}{\mathrm dt}}^{=H\,U_\mathrm S{(t,t^\prime)}}\,  e^{-iH_0t^\prime} \\
&= -e^{iH_0t}\,H_0\, U_\mathrm S(t,t^\prime) \, e^{-iH_0t^\prime} +  e^{iH_0t} \,(H_0 +V)\,U_\mathrm S (t,t^\prime)\, e^{-iH_0t^\prime} \\
&= e^{iH_0t}\, V \, U_{\mathrm S}(t,t^\prime) \, e^{-iH_0t^\prime} \\
&= e^{iH_0t}\, V \, e^{-iH_0t}\, e^{iH_0t} \,U_{\mathrm S}(t,t^\prime) \, e^{-iH_0t^\prime} \\
&= V_\mathrm I\, U_\mathrm I(t,t^\prime) \quad .
\end{align}
