Theorem: Let $\mathbf{A}$ be a locally small category and suppose $A, A' \in \mathbf{A}$. If $H_A \cong H_{A'}$ then $A \cong A'$. In other words, the Yoneda embedding $H_{\bullet}$ is injective on isomorphism classes of objects.

Proof:

$H_A \cong H_{A'}$ means we have a natural isomorphism $\alpha : H_A \rightarrow H_{A'}$. It has components $\alpha_X : H_A(X) \rightarrow H_{A'}(X) : \Hom(X, A) \rightarrow \Hom(X, A')$, which are also isomorphisms (in particular, the hom-sets are in bijection). Consider $\alpha_A : \Hom(A, A) \rightarrow \Hom(A, A')$ and denote $\ell = \alpha_A(1_A) : A \rightarrow A'$. We claim that this is the isomorphism desired. Indeed, first look at the naturality square of $\alpha$ with respect to the arrow $\ell$:

(the horizontal arrows in the opposite direction because $H_A$ is contravariant). Keep in mind the vertical arrows are bijections. The horizontal maps on both top and bottom are given by definition as right-composition by $\ell$, i.e. $p \mapsto p \circ \ell$. Since the vertical maps are bijections, we have an $r \in \Hom(A', A)$ mapping onto $1_{A'}$ via $\alpha_{A'}$. Evaluating the diagram at $r$ then gives:

Then commutativity says $\alpha_A(r \circ \ell) = \ell$. But $\ell$ was defined to be the image of $1_A$ under the bijection $\alpha_A$, so $r \circ \ell = 1_A$.

Let's look now at the naturality diagram with respect to this map $r$:

where we've swapped the vertical arrows with their inverses, for convenience. If we evaluate this diagram at $\ell$:

we get $\alpha_{A'}^{-1}(\ell \circ r) = r$. But $r$ was defined to be the preimage of $1_{A'}$ under $\alpha_{A'}$, so we have $\ell \circ r = 1_{A'}$.

Then, $\ell$ (or $r$) is the desired isomorphism between $A$ and $A'$.