Twisted Day convolution Has anyone studied a version of Day convolution for an enriched presheaf category $V^{A^{\mathrm{op}}}$ where the monoidal structure of $V$ is "twisted" on one side by an action of $A$?  I'm thinking of modifying the standard formula
$$(F \otimes G)(a) = \int^{b,c} F(b)\otimes G(c) \otimes A(b\otimes c,a)$$
to something like
$$(F \otimes G)(a) = \int^{b,c} F(b)\otimes G(c)^b \otimes A(b\otimes c,a)$$
where $G(c)^b$ denotes an action of $b\in A$ on $G(c)\in V$.
 A: The following is a proof that the twisted monoidal structure $\otimes^\rho$ is a convolution on $[A, V]$, where $A$ is a monoidal category, $V$ the cosmos on which $A$ is enriched, and $\rho\colon A\times V\to V$ a monoidal action (in the sense that $(v^a)^b \cong v^{a\cdot b}$, $v^i \cong v$ where $i$ is the monoidal identity of $A$, and $(v\otimes w)^a \cong v^a \otimes w^a$).
To make things work, I must assume that


*

*Every object $a$ of the monoidal structure $(A, \cdot, i)$ has a monoidal inverse $a^{-1}$;

*The functor $(-)^a \colon V\to V$ is cocontinuous.


This is somehow restricive, but since we know (coends, Prop. A.3) that (strong) promonoidal structure in $C$ correspond bijectively with convolution structures on $[A, V]$, there must exist a convolution product such that
$$
F\otimes^\rho G \cong \int^{xy} Fx \otimes Gy \otimes P_\rho(x,y;-)
$$
so that the action $(x,y)\mapsto (Gy)^x$ is "absorbed" into the promonoidal functor $P\colon A^\text{op}\times A^\text{op} \times A \to V$. 
I'm not able to find the $P_\rho$, but I'm able to find sufficient assumptions (those above) to ensure associativity. As I mentioned in the comment, I come up with a "twisted ninja Yoneda lemma" (coends, Prop. 2.1), and I learned something more about coend-nonsense, so thank you (-:
Let's start the coend-juggling:
\begin{align}
[(F \otimes^\rho G)\otimes^\rho H](a) &= \int^{bc} (F\otimes^\rho G)(b)\otimes (Hc)^b \otimes A(b\cdot c, a) \\
 & \cong \int^{bcxy} Fx \otimes (Gy)^x \otimes A(x\cdot y, b)\otimes (Hc)^b \otimes A(b\cdot c, a) \\
 & \cong \int^{bcxy} Fx \otimes (Gy)^x \otimes (Hc)^b \otimes A(x\cdot y, b)\otimes A(b\cdot c, a) \\
 & \cong \int^{cxy}Fx \otimes (Gy)^x \otimes \left( \int^b (Hc)^b \otimes A(x\cdot y, b)\otimes A(b\cdot c, a) \right) \\
 & \cong \int^{cxy}Fx \otimes (Gy)^x \otimes (Hc)^{x \cdot y}\otimes A((x\cdot y)\cdot c, a)
\end{align}
where I applied the ninja Yoneda lemma to the functor $S(b) = (Hc)^b \otimes A(b\cdot c, a)$ ($S$ results as the composition $b\overset{\Delta}\mapsto (b,b) \overset{\hat S}\mapsto (Hc)^b \otimes A(b\cdot c, a)$; this type of dependence is essential, and implicit in the definition). We now have to manipulate the other parenthesization of $F,G,H$: to obtain the same object, we need the assumptions of cocompleteness of the action and invertibility:
\begin{align}
[F \otimes^\rho (G\otimes^\rho H)](a) & \cong \int^{mn} Fm \otimes ((G\otimes^\rho H)(n))^m \otimes A(m\cdot n, a) \\
& \cong \int^{mnzt} Fm \otimes \Big( Gz\otimes (Ht)^z \otimes A(z\cdot t, n)\Big)^m \otimes A(m\cdot n, a) \\
& \cong \int^{mnzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes A(z\cdot t, n)^m \otimes A(m\cdot n, a) \\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes \left( \int^n A(z\cdot t, n)^m \otimes A(m\cdot n, a) \right)\\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes \left( \int^n A(z\cdot t, n) \otimes A(m\cdot n, a)^{m^{-1}} \right)^m\\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes \Big( A(m\cdot (z\cdot t), a)^{m^{-1}} \Big)^m\\
& \cong \int^{mzt} Fm \otimes (Gz)^m \otimes (Ht)^{z\cdot m} \otimes A(m\cdot (z\cdot t), a)
\end{align}
where I applied the assumptions, and the ninja Yoneda lemma.
The same manipulations show that the presheaf $\hom(i,-)$ is the monoidal unit.
\begin{align}
F \otimes^\rho I & = \int^{bc} Fb \otimes A(i,c)^b \otimes A(b\cdot c ,a ) \\
& \cong \int^b Fb \otimes \left( \int^c A(i,c) \otimes A(b\cdot c, a)^{b^{-1}} \right)^b \\
&\cong \cdots
\end{align}
