I'm following along Wikipedia page(https://en.wikipedia.org/wiki/Universal_property) on universal property, and this seems it should be trivial, but I couldn't finish the proof.
The definition I am working with: 
The problem
So I want to prove that if $(A, u)$ and $(A', u')$ are universal morphism from $X$ to $F$, then there exists unique isomorphism $k: A \rightarrow A'$ such that $u' = F(k) \circ u$.
My attempt:
For any $B \in C$ and $f: X \rightarrow F(B)$, there exists unique $h: A \rightarrow B$ such that $f = h \circ u$.
Letting $B = A'$ and $f = u'$, we see that there exists unique $k: A \rightarrow A'$ such that $Fk \circ u = u'$.
This shows existence and uniqueness. We only need that $k$ is isomorphism, but this is the part I cannot prove.
Similarly, there exists unique $k': A' \rightarrow A$ such that $Fk' \circ u' = u$.
Plugging into each equation, we get
$$Fk \circ Fk' \circ u' = F(k \circ k') \circ u' = u'$$
$$Fk' \circ Fk \circ u = F(k' \circ k) \circ u = u$$
but this of course doesn't imply $F(k \circ k') = id$, so it seems useless.
Thanks for your help!
You've derived the equation $F(k'\circ k)\circ u=u$. So $k'\circ k\colon A\to A$, but since $(A,u)$ is a universal morphism, by definition there is a unique morphism $h\colon A\to A$ such that $F(h)\circ u=u$. (This is the condition you get when you apply the universal property to the universal morphism itself.) But $h=1_A$ also works, so by uniqueness, $k'\circ k=1_A$.
Reversing the roles, as you've already done, shows $k\circ k'=1_{A'}$, so $k$ and $k'$ are isomorphisms.