Let $G$ be a linear algebraic group, let $X$ be a $G$-variety (for simplicity, let $X$ be a complex affine variety, with associated structure sheaf $\mathcal O_X$, regarded as a sheaf of $\mathcal O_X$-modules) with $G$-action $\sigma: G \times X \to X$ and projection $p: G \times X \to X$. Then $\sigma^* \mathcal O_X$ and $p^* \mathcal O_X$ are sheaves on $G \times X$ (considered as the product of two varieties). Since $$ \sigma^* \mathcal O_X = \mathcal O_{G \times X} \otimes_{\sigma^{-1} \mathcal O_X} \sigma^{-1} \mathcal O_X \cong \mathcal O_{G \times X} $$ and a similar expression holds for $p^* \mathcal O_X$, there is a natural isomorphism $\sigma^* \mathcal O_X \cong p^* \mathcal O_X$. I would like to show that this satisfies the cocycle condition in the definition of $G$-equivariant sheaves, namely $$ p_{23}^* \phi \circ (1_G \times \sigma)^* \phi = (m \times 1_X)^* \phi,$$ where $p_{23} : G \times G \times X \to G \times X$ is projection along the first factor and $m$ is the group multiplication on $G$.
Here is what I have tried. Evaluate (I think it makes sense to 'evaluate' these morphisms, considered as morphisms on $\text{Sh}(G \times G \times X)$) the left-hand side on the sheaf $(1_G \times \sigma)^* \sigma^* \mathcal O_X$: $$ \begin{aligned} p_{23}^* \phi \circ (1_G \times \sigma)^* \phi \left( (1_G \times \sigma)^* \sigma^* \mathcal O_X \right) &= p_{23}^* \phi \circ (1_G \times \sigma)^* \left( \phi (\sigma^* \mathcal O_X) \right) \\ &= p_{23}^* \phi \circ (1_G \times \sigma)^* \left( p^* \mathcal O_X \right) \\ &= p_{23}^* \phi \circ (1_G \times \sigma)^* \mathcal O_{G \times X} \\ &= p_{23}^* \phi \left( \mathcal O_{G \times G \times X} \otimes_{(1_G \times \sigma)^{-1} \mathcal O_{G \times X}} (1_G \times \sigma)^{-1} \mathcal O_{G \times X} \right) \\ &= p_{23}^* \phi (\mathcal O_{G \times G \times X}). \end{aligned} $$ The first equality I justify since pullback is functorial, and the rest is just unpacking definitions. What I am confused on is how to continue - the thing I have ended up with seems resistant to further manipulation. Moreover, evaluation of the right-hand side also confuses me - the expression $$ (m \times 1_X)^* \phi \left( (1_G \times \sigma)^* \sigma^* \mathcal O_X \right)$$ looks like nonsense to me.
I understand vaguely how the cocycle condition reflects the group associativity, though this I am still not sure on (I have the example of $G = \mathbb{C}^\times, X = \mathbb{V}(x)$ in mind), but I do not see how to concretely show $\mathcal O_X$ is an equivariant sheaf. What am I missing in all of this?
The calculation is very straightforward once it is observed that there are three commutative diagrams projecting from $G \times G \times X$ to $X$: $$ \require{AMScd} \begin{CD} G \times G \times X @>m \times 1>> G \times X\\ @VV1 \times aV @VVaV\\ G \times X @>a>> X \end{CD}, \require{AMScd} \begin{CD} G \times G \times X @>p_{23}>> G \times X\\ @VV1 \times aV @VVaV\\ G \times X @>p>> X \end{CD}, \require{AMScd} \begin{CD} G \times G \times X @>p_{23}>> G \times X\\ @VVm \times 1V @VVpV\\ G \times X @>p>> X \end{CD}. $$
In this way, there are three sets of equalities that hold (regardless of if $\mathcal F$ is equivariant or not): $$ \begin{aligned} (1 \times a)^* a^* \mathcal F &= (m \times 1)^* a^* \mathcal F, \\ (1 \times a)^* p^* \mathcal F &= p_{23}^* a^* \mathcal F, \\ (m \times 1)^* p^* \mathcal F &= p_{23}^* p^* \mathcal F. \\ \end{aligned} $$ Using these three equalities, the calculation can be fixed.