I just wanted to verify if my proof is correct as I have to present this proof in my next class.
Consider the comm. diagram drawn above. We can verify that it commutes due to the categorical definition of a free group. Now $\exists$ unique group homomorphisms $\overline \varphi, \overline \varphi'$ such that $\overline \varphi \circ \overline \varphi' : V \to V$ is an automorphism on $V$. However, since $\overline \varphi, \overline \varphi'$ are unique for fixed $U \& V$, their product must also be unique. Now obviously $id_V : V \to V, x \to x \ \ \forall x \in V$ is an automorphism, we must have that $\overline \varphi \circ \overline \varphi' = id_V$. A similar argument can be made for $U$. Hence, there exists unique homomorphisms $\overline \varphi , \overline \varphi'$ such that $\overline \varphi \circ \overline \varphi' = id_V \text{and } \overline \varphi' \circ \overline \varphi = id_U $ Hence, we have that $U \cong V$