When is an object-preserving autoequivalence isomorphic to the identity? Consider a triangulated category $\mathcal{T}$ and an exact autoequivalence $\Phi:\mathcal{T}\rightarrow \mathcal{T}$ such that $\Phi(F)=F$ for any object $F$ in $\mathcal{T}$, when could we say that $\Phi$ is isomorphic to the identity functor?
 A: As a warm-up, consider an equivalence $F\colon\text{Set}\to\text{Set}$ that is the identity on objects.  For each set $X$ and $x\in X$ we have a map $i_x\colon1=\{0\}\to X$ sending $0$ to $x$.  This gives $F(i_x)\colon 1\to X$, which has the form $i_{\alpha_X(x)}$ for some element $\alpha_X(x)\in X$.  As $F$ is an equivalence, we see that $\alpha_X\colon X\to X$ is a bijection.  For any $f\colon X\to Y$, we can apply functoriality to the relation $f\circ i_x=i_{f(x)}$ to get $(Ff)\circ i_{\alpha_X(x)}=i_{\alpha_Y(f(x))}$.  Using this we get $Ff=\alpha_Y\circ f\circ\alpha_X^{-1}$.  Thus, the maps $\alpha_X$  give a natural isomorphism $1\to F$.
Now consider an additive equivalence $F\colon\text{Ab}\to\text{Ab}$ that is the identity on objects.  This again gives isomorphisms
$$ \alpha_A=F\colon A=\text{Ab}(\mathbb{Z},A) \to 
    \text{Ab}(F\mathbb{Z},FA) = \text{Ab}(\mathbb{Z},A) = A.
$$
Essentially the same argument shows again that for $f\colon A\to B$ we have $Ff=\alpha_B\circ f\circ\alpha_A^{-1}$, so $F$ is isomorphic to the identity.
Now consider an exact equivalence $F\colon D(\text{Ab})\to D(\text{Ab})$ that is the identity on objects.  Exactness includes the condition that for $f\colon X\to Y$ we have $F(\Sigma f)=\Sigma(Ff)\colon\Sigma X\to\Sigma Y$.  We can identify $H_n(X)$ with $D(\text{Ab})(\Sigma^n\mathbb{Z},X)$ and thus obtain automorphisms $\alpha_X\colon H_*(X)\to H_*(X)$ with $H_*(Ff)=\alpha_Y\circ H_*(f)\circ\alpha_X^{-1}$ for all $f\colon X\to Y$.  As $F$ is an equivalence it must preserve all coproducts, and using this we see that $\alpha_{\bigoplus_iX_i}=\bigoplus_i\alpha_{X_i}$.  A standard argument shows that any object in $D(\text{Ab})$ can be expressed as a direct sum of two-term complexes, where the unique nontrivial differential is an injective homomorphism of free abelian groups.  This means that the complex is essentially a projective resolution of the unique homology group, so every automorphism of the homology lifts (but not uniquely) to an endomorphism of the complex.  Using this, we see that we can choose a chain equivalence $\beta_X\colon X\to X$ for each $X$, so that $H_*(\beta_X)=\alpha_X$.  This means that $F$ is isomorphic to the functor $F'(f)=\beta_Y^{-1}\circ F(f)\circ\alpha_X$ which satisfies $H_*(F'(f))=H_*(f)$.  Now put
$$ \mathcal{I}(X,Y) = \{u\in D(\text{Ab})(X,Y) : H_*(u)=0\} $$
One can check (using direct sum splittings as before) that if $v\in\mathcal{I}(Y,Z)$ and $u\in\mathcal{I}(X,Y)$ then $v\circ u=0$.  In other words, $\mathcal{I}$ is a square-zero categorical ideal in $D(\text{Ab})$.  We have $F'(f)=f+D(f)$ for some $D\colon D(\text{Ab})(X,Y)\to\mathcal{I}(X,Y)$. Using functoriality and the square-zero property we find that $D$ is a derivation in the sense that $D(g\circ f)=D(g)\circ f+g\circ D(f)$.  We also find that $F'$ is isomorphic to the identity iff there is a system of elements $P_X\in\mathcal{I}(X,X)$ satisfying $D(f)=P_Y\circ f-f\circ P_X$ for all $f\colon X\to Y$.  I do not know whether that is always satisfied.
I expect that almost any other triangulated category will give a harder problem.
