Let $M=\mathbb C^g/ \Gamma$ be a complex tori and $E$ a be a holomorphic vector bundle of rank $r$ over $M$. Then $E$ is characterised by factor of automorphy, i.e. a holomorphic map $J:\Gamma\times\mathbb C^g\to GL(r,\mathbb C)$ such that $J(\gamma'\gamma,x)=J(\gamma',\gamma x)J(\gamma,x)$. If $f:M\to M$ is a holomorphic diffeomorphism of $M$, $f^\*(E)$ is the pull-back bundle. Then can we deduce that $f^\*(E)$ is given by a factor of automorphy $J\_f(\gamma,x)=J(\gamma,f(x))$ ?



Pull-back of factor of automorphy
I think every holomorphic map $f:\mathbb{C}^g/\Gamma\to \mathbb{C}^g/\Gamma$ lifts to a $\Gamma$-equivariant holomorphic map $\tilde{f}:\mathbb{C}^g\to \mathbb{C}^g$ (indeed, every holomorphic map is a composition of a homomorphism, that lifts, with a translation, that lifts too).


Hence a factor of automorphy giving $f^\*E$ is obtained as the pull-back of $J$ by $\tilde{f}$.


I don't think that you need $f$ to be biholomorphic for this to be true.


EDIT: $\Gamma$ -equivariance shall be understood in the following sense. There exists a group morphism $\varphi:\Gamma\to\Gamma$ such that $\tilde{f}(\gamma\cdot x)=\varphi(\gamma)\cdot\tilde{f}(x)$. Then the pulled-back factor of automorphy shall be $J\big(\varphi(\gamma),\tilde{f}(x)\big)$.


EDIT2: here $\cdot$ means $+$.


