Picard group of UFD is trivial

The point is that if $\varphi\in\operatorname{Hom}_R(L,R)$ is nonzero, then it is injective so it gives an isomorphism from $L$ to $\varphi(L)$, and $\varphi(L)\cong R$ since it is a principal ideal.

Here is one way you can see that $\varphi$ is injective. Note that if $K$ is the fraction field of $R$ then $K\otimes L\cong K$ (since $L$ is locally free of rank $1$) and $\varphi$ induces a homomorphism $\varphi\otimes K:L\otimes K\to R\otimes K$. Since $L$ is torsion free, it embeds in $L\otimes K$, so if $\varphi$ is nonzero then so is $\varphi\otimes K$. But since $L\otimes K$ and $R\otimes K$ are both $1$-dimensional vector spaces, this means $\varphi\otimes K$ is injective, and hence so is $\varphi$.