Let's call our map $f:X \to Y$, the kernel $K$, and the kernel pair $P$. As you said, the natural map from the kernel to the kernel pair comes from a map of pullback squares. If this has an inverse, then chasing this diagram shows that the map $P \to Y$ factors through zero. But the map $X \to Y$ factors through $P$, since $P$ is the pullback of $X \to Y \leftarrow X$ and there is also a commutative square
X -> X
X -> Y
where the top and left-hand maps are the identity.
Thus the map $X \to Y$ is zero. Your condition only holds in categories equivalent to the terminal category!