When should we consider "reverse Heisenberg" evolution of operators? In Quantum Mechanics, the Heisenberg evolution of an observable $\hat{o}$ is defined as
$$ \hat{o}(t) = U(t,0)^{\dagger} \hat{o} U(t,0) $$
where $U(t,0)$ is the unitary time-evolution operator from time $0$ to time $t$. This satisfies the Heisenberg equations of motion
$$ i\hbar \frac{d}{dt} \hat{o}(t) = [\hat{o}(t),H(t)],$$
But is there a standard name for the "reverse Heisenberg evolution"
$$ \hat{o}_R(t) = U(t,0) \hat{o} U(t,0)^{\dagger} $$
which satifies the differential equation
$$ i\hbar \frac{d}{dt} \hat{o}_R(t) = [H(t),\hat{o}_R(t)],$$
and in which circumstances should one consider it?
It came up because I was thinking about a state $|\psi\rangle$ which is defined to be the (unique, say) eigenstate of some observable $\hat{o}$ with eigenvalue $\lambda$. Then we see that the time-evolved state $|\psi(t)\rangle = U(t,0) |\psi\rangle$ can be characterized as the eigenstate of the operator $\hat{o}_R(t)$.
 A: 
It came up because I was thinking about a state $|\psi\rangle$ which is defined to be the (unique, say) eigenstate of some observable $\hat{o}$ with eigenvalue $\lambda$. Then we see that the time-evolved state $|\psi(t)\rangle = U(t,0) |\psi\rangle$ can be characterized as the eigenstate of the operator $\hat{o}_R(t)$

This property is necessary for the derivation of the path integral in quantum mechanics and QFT. In the context I saw it, it was used in reverse, but it's the same idea. We had operators which evolve normally
$$X(t) = U^\dagger (t)X U(t)$$
$$P(t) = U^\dagger (t)P U(t)$$
but states were then defined with backwards evolution:
$$|x,t\rangle \equiv U^\dagger (t)|x\rangle$$
So that for all time, they remain eigenstates:
$$X(t)|x,t\rangle=U^\dagger (t)X U(t) U^\dagger(t)|x\rangle$$
$$=U^\dagger (t)x |x\rangle$$
$$=x |x,t\rangle$$
(and same with $P$)
If you want to read more, I learned this from Weinberg's The Quantum Theory of Fields, the part he does this is in Chapter 9.1, page 379. His derivation is for Quantum Mechanics but with n degrees of freedom. Taking the $n\to \infty$ would give the QFT path integral.
You asked if this evolution has a name; I don't know of any special name for it. Reverse evolution seems to get the point right.
