Let $X$ be a smooth variety (over $\mathbb{C}$) and $\Delta: X \rightarrow X \times X$ be the diagonal embedding and $p\_1: X\times X\rightarrow X, ~p\_2: X\times X\rightarrow X$ be the projections to the first and second components. Let $E$ be a finite dimensional vector bundle on $X$. We define
$$
E\_{\Delta}:=\Delta\_\*E
$$
to be a sheaf on $X\times X$. In particular we have $\mathcal{O}\_{\Delta}:=\Delta\_\*\mathcal{O}$. We have the sheaf Ext functor $\mathcal{Ext}$ on $X\times X$.


$\bf{My~question}$ is: do we have the isomorphism
$$
{Rp\_1}\_\*\mathcal{Ext}\_{X\times X}^{\bullet}(E\_{\Delta},E\_{\Delta})\cong 
{Rp\_1}\_\*\mathcal{Ext}\_{X\times X}^{\bullet}(\mathcal{O}\_{\Delta},\mathcal{O}\_{\Delta})\otimes \mathcal{End}(E) ?
$$


I know that when $X$ is affine, they are isomorphic since $E$ is then free. But does it also hold in the general case?



Do we have the following isomorphism for $\mathcal{Ext}$?
Note that $E \cong \Delta^\*p\_1^\*(E)$ since $p\_1\circ\Delta = 1\_X$. Therefore by the projection formula
$\Delta\_\*E = \Delta\_\*\Delta^\*p\_1^\*(E) = p\_1^\*(E)\otimes\Delta\_\*O\_X$, hence
\begin{align\*}
\mathcal{Ext}(\Delta\_\*E,\Delta\_\*E) & 
= \mathcal{Ext}(p\_1^\*(E)\otimes\Delta\_\*O\_X,p\_1^\*(E)\otimes\Delta\_\*O\_X) \\
& = p\_1^\*(E^\*)\otimes p\_1^\*(E)\otimes\mathcal{Ext}(\Delta\_\*O\_X,\Delta\_\*O\_X) = \\
& = p\_1^\*\mathcal{End}(E)\otimes\mathcal{Ext}(\Delta\_\*O\_X,\Delta\_\*O\_X),
\end{align\*}
hence by the projection formula
\begin{align\*}
Rp\_{1\*}\mathcal{Ext}(\Delta\_\*E,\Delta\_\*E) 
& = Rp\_{1\*}(p\_1^\*\mathcal{End}(E)\otimes\mathcal{Ext}(\Delta\_\*O\_X,\Delta\_\*O\_X)) \\ 
& = \mathcal{End}(E)\otimes Rp\_{1\*}\mathcal{Ext}(\Delta\_\*O\_X,\Delta\_\*O\_X)).
\end{align\*}


