Let $e_1, \cdots, e_n$ be a local basis of $E$ so that $\
abla e_i (x) = 0$ at $x\in M$. Using $1.$, the dual basis $e_1^*, \cdots, e_n^*$ also has $\
abla e_i^* (x) = 0$. Using this basis, we have locally $$\text{Id} = \sum_{i=1}^n e_i^* \otimes e_i,$$ thus $\
abla \text{Id}(x) = 0$ by $2.$. Since $x\in M$ is arbitrary, we have $\
abla \text{Id} = 0$.