Consider the functors $H_A, X:\mathscr A^{op}\to\mathbf {Set}$.
There's the following corollary of the Yoneda lemma:
There's a bijection $$\{ \text{natural isomorphisms } \alpha: H_A\to X\}\leftrightarrow \{\text{elements }x\in X(A) \text{ satisfying } (\ast)\}$$ ($\ast$) For all $B\in \mathscr A$ and for all $x\in X(B)$ there's a unique $\overline x:B\to A$ such that $X(\overline x)(u)=x$.
I read the proof in Leinster's textbook (Corollary 4.3.2), but it turned out to be (relatively) much shorter than the proof that I came up with. He uses one "reduction" that I'm not sure if I understand. Before getting to his reduction, here's my straightforward proof (just sketch, for the details see the very bottom of the question):
We know from the Yoneda lemma that there is a bijection $$[\mathscr A^{op},\mathbf{Set}](H_A,X)\simeq X(A)$$ given from left to right by $\alpha\mapsto \hat\alpha$ where $\hat\alpha=\alpha_A(1_A)$, and from right to left by $u\mapsto \widetilde u$ where $\widetilde u_B:H_A(B)\to X(B)$ is given by $f\mapsto X(f)(u)$. So it remains to prove two things:
(1) if $\alpha$ is a natural isomorphism, then $\hat \alpha$ satisfies ($\ast$) with $u=\hat\alpha$;
(2) if $u\in X(A)$ satisfies ($\ast$), then $\widetilde u$ is a natural isomorphism (it's already known that $\widetilde u$ is a natural transformation, so in fact it suffices to show that $\widetilde u_B$ is bijective for all $B$).
Instead of proving (1) and (2), Leinster says that it's enough to prove that $\widetilde u$ is an isomorphism $\iff$ ($\ast$) holds. The right-to-left implication is exactly (2) above. But the left-to-right implication doesn't look like (1). (It's a particular case of (1) with $\alpha=\widetilde u$.) Leinster's implication from left to right is completely obvious, but (1) is not so obvious; in particular, I needed to use the naturality of $\alpha$ to prove (1), whereas the naturality of $\widetilde u$ is not used in Leinster's proof of the left-to-right implication.
So my question is why can proving (1) and (2) be reduced to proving the equivalence that Leinster suggested? In other words, why is it enough to prove (2) and the particular case of (1) with $\alpha=\tilde u$?
Not directly realted to the question, but maybe someone will find this useful in the future.
Proof of (1):
Suppose $\alpha$ is a natural isomorphism. In particular, the components $$\alpha_B: H_A(B)\to X(B)\\f\mapsto \alpha_B(f)$$ are bijective. This means that for all $y\in X(B)$ there is a unique $\overline y:B\to A$ such that $\alpha_B(\overline y)=y$. But we need to prove that for all $y\in X(B)$ there is a unique $\overline y:B\to A$ such that $X(\overline y)(\hat \alpha)=y$. If we write the naturality conditions for $\alpha$ w.r.t. the arrow $\overline y$, we'll get that $X(\overline y)(\hat \alpha)=\alpha_B(\overline y)$. This completes the proof of (1).
Proof of (2):
The condition ($\ast$) says exactly that $\widetilde u_B$ is bijective for all $B$.