Fourier-Mukai functors and autoequivalence groups of $G$-equivariant derived categories I have a few questions about $G$-equivariant derived categories. For my question, I'm assuming $G$ is cyclic. Also, in my case $G$ does not act on $X$, only on $D^b(X)$.
Q1: Orlov's Representability Theorem states that any fully faithful functor between derived categories of smooth projective varieties is Fourier-Mukai. Is there a version of this theorem for $G$-equivariant derived categories? In other words, if $X$ and $X'$ are smooth projective varieties and $\Phi^G : D^b(X)^G \simeq D^b(X')^G$ is fully faithful (actually, for my purposes $\Phi^G$ can be an equivalence), is it true that $\Phi^G$ is Fourier-Mukai? If the notion of a Fourier-Mukai functor even makes sense for $G$-equivariant derived categories... By David Ploog's paper Equivariant autoequivalences for finite group actions the concept does seem to extend.
Q2: If $X$ is Fano or general type, we know by Bondal-Orlov that $\mathrm{Aut}(D^b(X)) = \mathrm{Aut}(X) \ltimes (\mathrm{Pic}(X) \oplus \mathbb{Z})$. Can we conclude anything about $\mathrm{Aut}(D^b(X)^G)$ (of course if $G$ acts appropriately on $D^b(X)$)? Again, David Ploog's paper seems useful for this since he compares $\mathrm{Aut}(D^b(X)^G)$ with $\mathrm{Aut}(D^b(X))^G$.
 A: Since no one has yet chimed in, I'll try to at least partially address your questions.

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*The notion of a Fourier-Mukai functor does extend. If you have a $G$-variety $X$ and a $H$-variety $Y$, then the pullback functor $\pi^*:D^b_G(X) \to D^b_{G \times H}(X \times Y)$ exists and is in fact essentially just the normal pullback. Similarly, the pushforward functor exists (with one minor modification, you need to take $G$-invariants). More details about this can be found in  this paper. Now whether or not every fully faithful functor with an adjoint is such a functor, it seems the answer is not completely clear. Kawamata has a paper here which addresses the question in a particular case, but its not clear that its been done in full generality. However it could turn out to be easy, and so no one has really written it down.


*It will very much depend on the group action in question. For example in this paper that authors give a handful of examples of an abelian surface with group action such that the quotient is $\mathbb{P}^2$. Thus it would follow that the automorphism group of $D^b(A)$ is not really so related to the automorphism group of $D^b_G(A) \cong D^b(\mathbb{P}^2)$.
Hope this helps.
