Let $G$ be a finite group and $H \subset G$ a normal subgroup. Consider $G$, $H$, and $X=G/H$ as affine algebraic groups over some algebraically closed base field $k$.

I hear that there is an isomorphism of stacks $[X/G] \cong [pt/H]$.

I have the following question:

To give a sheaf (of vector spaces) on the stack $[X/G]$ is the same as giving a $G$-equivariant sheaf on $X$. By the isomorphism above, it is the same as giving a vector space with an $H$ action.

What is this functor taking $G$-equivariant modules over the ring $k[G/H]=k[G]^H$ to vector spaces with $H$ actions?

For example, what happens to the $G$-equivariant $k[G/H]$-module $M=k[G]$?

*** Edit to more general situation

The answers are getting stuck in a very basic situation, I want to think of a more general situation.

Suppose that $G$ is an affine algebraic group over an algebraically closed subfield, $H$ a normal subgroup, and $X$ an affine $G$-variety with action factoring through $G/H$. Suppose that $G/H$ acts properly and freely on $X$. The stack $[X/(G/H)]$ is representable by a scheme $X/(G/H)$.

Question 1: Do we still have $[X/G] \cong [(X/(G/H))/H]$?

If so, Question 2: For any $G$-equivariant sheaf $\mathcal{M}$ on the space $X$, by descent for Cartesian sheaves $\mathcal{M}(X/(G/H))$ is computed by the kernel of the diagram,

$$\mathcal{M}(X \times_{[X/G]} X/(G/H)) \rightarrow \mathcal{M}(X \times_{[X/G]} X/(G/H) \times_{X/(G/H)} X \times_{[X/G]} X/(G/H))$$

What is this functor?

Example: When $H = e$, this equalizer takes the difference between the action and projection pull-backs yielding the functor of invariants.