Let $X$ be a projective variety over $\mathbb{C}$, denote by $D^b(X)$ the bounded derived category of coherent sheaves on $X$. Suppose we have a Fourier-Mukai functor $\Phi_{X\rightarrow X}^\mathcal{P}:D^b(X)\rightarrow D^b(X)$ being an auto-equivalence on $D^b(X)$, and further assume that $\Phi_{X\rightarrow X}^\mathcal{P}$ acts identically on objects, then is it possible that $\Phi_{X\rightarrow X}^\mathcal{P}$ fails to be the identity functor? How to find a simple example to illustrate this?

If $X$ is *smooth* and projective, then any such FM functor is in fact naturally isomorphic to the identity functor. This follows immediately from Corollary 5.23 of Huybrechts' book on Fourier-Mukai tranforms. Briefly, the idea is that the hypotheses ensure that $\mathcal{P}$ is a quasi-isomorphic to a sheaf on $X\times X$ that's supported set-theoretically on the diagonal and moreover is flat over $X$ via either projection map. One then argues that $\mathcal{P}$ is of the form $\mathcal{O}_\sigma\otimes L$, where $\mathcal{O}_\sigma$ is the structure sheaf of the graph of an automorphism of $X$ and $L$ is a line bundle pulled back from $X$.

When $X$ is not smooth, I'm not completely sure what happens. If $P$ is in fact a perfect complex on $X\times X$, then this reasoning still goes through. Otherwise, one needs to worry about the difference between $D^b(X)$ and $D_{perf}(X)$, the derived category of perfect complexes on $X$.