Suppose that $A$ is an abelian group and $B$ is a subgroup of $A$. Then there are usually plenty of homomorphisms (actually they must be automorphisms) $\theta:A \to A$ which restrict to the identity on $B$ and induce the identity on $A/B$. You can think of them as skew transformations, acting along the cosets of $B$. They can all be uniquely expressed in the form $\theta = \mathrm{id}_A + T\circ \pi$ where $\pi$ is the projection $A \to A/B$ and $T : A/B \to B$ is an arbitrary homomorphism.
But what if $R$ is a ring and $I$ is an ideal in $R$? Then you are hard-pressed to find ring homomorphisms $\theta$ which restrict to the identity on $I$ and induce the identity on $R/I$. Of course you can take the zero multiplication on $R$ (assuming your worldview is accepting of nonunital rings) which brings us back to the setting of abelian groups, so at least there are some examples, but I'm not sure if there are any more compelling ones.
My recollection is that there are no such maps, apart from the identity, in the context of C*-algebras and (closed, $*$-invariant) ideals, but I can't recall at the momement whether this is an analytic fact (say using the uniqueness of the norm on the quotient or something) or if it just follows from some purely algebraic considerations which I am overlooking.
Here's a simple example. Let $R=\mathbb{Z}[x,y]/(xy,y^2)$ and let $f:R\to R$ be given by $f(x)=x+y$ and $f(y)=y$. This restricts to the identity on the ideal $I=(y)$ and induces the identity $R/I\to R/I$, but is not the identity.
One simple obstruction is that if a (left) ideal $I\subseteq R$ contains an element $a$ that is not a (right) zero divisor, then any homomorphism $f:R\to R$ that restricts to the identity on $I$ must be the identity. Indeed, for each $r\in R$, $ra\in I$ so $ra=f(ra)=f(r)f(a)=f(r)a$, which implies $r=f(r)$ since $a$ is not a zero divisor. A bit more generally, the same conclusion holds by a similar argument if $I$ is a faithful $R$-module.
More strongly, suppose $f:R\to R$ restricts to the identity on $I$ and induces the identity $R/I\to R/I$. Then for each $r\in R$, $f(r)-r\in I$ (since $f$ induces the identity $R/I\to R/I$), but the calculation above shows that $(f(r)-r)I=0$. So, if $f$ is not the identity, there must be a nonzero element of $I$ that annihilates $I$. This is impossible for a $*$-ideal $I$ in a $C^*$-algebra, since if $a\in I$ then $a^*\in I$ but $aa^*=0$ implies $a=0$.