Evolution operator for "blending" a pair of eigenstates Consider a measurement operator ("observable") $\hat O$ which has ("a spectrum of") only two distinct eigenvectors; formally 
$$\hat O |\bullet\rangle := r_{\bullet}~|\bullet\rangle, \qquad \hat O |\circ\rangle := r_{\circ}~|\circ\rangle, \qquad \langle \bullet |\circ\rangle = 0, $$
where the eigenvalues $r_{\bullet}$ and $r_{\circ}$ are (suitable) real numbers.
Is there a time-dependent unitary operator $\hat U[~t~]$ such that 
$$ \frac{|\bullet\rangle + |\circ\rangle}{\sqrt 2} = \hat U[~t_{\text{blend}}~] ~ |\bullet\rangle $$ 
?
If so, can this operator be expressed in the form  $\hat U[~t~] := \text{Exp}[~\frac{-i~t}{\hbar}~\hat H[~t~]~]$ ?
If so, can the corresponding operator $\hat H[~t~]$ be expressed in terms of the given operator $\hat O$, or its eigenvectors or eigenvalues, and, as necessary, the coordinate value $t_{\text{blend}}$ and the coordinate variable $t$?

Edit
(along with correcting the above "evolution equation" stated within the bra-ket formalism):    
Within the density matrix formalism it could be asked correspondingly whether there is a time-dependent unitary operator $\hat U[~t~]$ such that 
$$ \sqrt{\frac{1}{2}} \left( \array{ |\bullet\rangle\langle\bullet | & |\bullet\rangle\langle\circ | \cr |\circ\rangle\langle\bullet | & |\circ\rangle\langle\circ | } \right) = \hat U[~t_{\text{blend}}~] ~ \left( \begin{array}{cc} |\bullet\rangle\langle\bullet | & 0 \\ 0 & |\circ\rangle\langle\circ | \end{array} \right) ~ \hat U[~t_{\text{blend}}~]^{\dagger}.$$
 A: OK firstly operators act on kets as vectors, so for some unitary operator $\hat{U}$, the action of $\hat{U}$ sends $|\psi\rangle\rightarrow \hat{U}|\psi\rangle$ not $\hat{U}|\psi\rangle\hat{U}^\dagger$, which is normally not a well defined expression.
For the rest of your question, since we are only dealing with a two dimensional space (or at least only a two dimensional subspace of some larger space) it is clearest to use matrix notation. So
\begin{equation}
\hat{U}(t) = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)\quad 
|\bullet\rangle = \left(\begin{array}{c}1\\0\end{array}\right)\quad 
|\circ\rangle = \left(\begin{array}{c}0\\1\end{array}\right)
\end{equation}
In this notation your blending condition becomes \begin{equation}
\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1 \end{array}\right) = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)\left(\begin{array}{c}1\\0\end{array}\right) = \left(\begin{array}{c}a\\c\end{array}\right)
\end{equation}
The requirement that $\hat{U}$ be unitary then fixes $b$ and $d$ up to a sign, so it is easy to work out that\begin{equation}
\hat{U} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc}1& \pm 1\\1&\mp 1\end{array}\right)
\end{equation}
If we put the minus sign in the top right (the case where it is in the bottom right differs only by a change of basis) then $\hat{U}$ has a standard representation in terms of Pauli matrices\begin{equation}
\hat{U} = \exp\left(-\imath\frac{\pi}{4}\sigma^{\,y}\right)
\end{equation} 
