Viewing functors $1\to\mathcal A$ as objects of $\mathcal A$, a left adjoint to $\mathcal A\to 1$ is exactly an initial object of $\mathcal A$.
Let $Ob(1)=\{\star\}$, let $F:1\to \mathcal A$ be the functor that assigns to the object of $1$ the initial object of $\mathcal A$ and let $G:\mathcal A\to 1$ be the constant functor. The above means at least that there is a bijection $$\mathcal A (F(\star),A)\simeq1(\star,G(A))$$ where $A\in Ob(\mathcal A)$; or, since $F(\star)$ is the initial object in $\mathcal A$, in symbols $F(\star)=I$, and since $G(A)=\star$, there is a bijection $$\mathcal A(I,A)\simeq 1(\star,\star)$$ (and the compositions are identities).
I couldn't figure out how to assign to an arrow $I\to A$ and arrow $\star\to \star$ (or vice versa). How to do that? How to use that $I$ is initial?
Note that $1(\star, G(A)) = 1(\star, \star) = \{ \mathrm{id}_{\star} \}$, so:
If there is a bijection $\mathcal{A}(F(\star), A) \cong 1(\star, G(A))$ for each $A \in \mathcal{A}$, then $\mathcal{A}(F(\star), A)$ has a unique element. This unique element is precisely the morphism ${!}_A : F(\star) \to A$ witnessing that $F(\star)$ is initial.
Conversely, if $I$ is initial, then $\mathcal{A}(I,A)$ has a unique element ${!}_A : I \to A$, and so there is automatically a bijection $\mathcal{A}(I,A) \cong 1(\star, \star)$ since both of these sets have one element.