Is the underlying vector space of the minimal model of an $A_{\infty}$-algebra canonical? On the page 4 of these notes it is stated that an $A_{\infty}$-algebra $A$ is necessarily is quasi-isomorphic to an $A_{\infty}$-algebra $HA$ with trivial differential. Moreover, $HA$ is unique up to a non-unique $A_{\infty}$-isomorphism. My question, is the morphism between the underlying vector spaces induced by this non-unique isomorphism unique?
 A: You're essentially asking whether an $A_\infty$-algebra with zero differential has nontrivial auto-isotopies (where an isotopy is an $\infty$-morphism whose linear part is an isomorphism). Indeed, if there is a nontrivial one, then you can compose your isomorphism with it to get a different one; conversely if you have two isomorphisms, then you take $fg^{-1}$ to get a nontrivial automorphism.
Nontrivial automorphisms exist in general, so the answer to your question is no, there is no unicity. For a simple example, consider $A = HA = \mathbb{R}[x]/(x^2)$ (for some $x$ of degree $n > 0$) with the obvious algebra structure. Then there is a nontrivial automorphism given by $f: A \to A$, $f(x) = 2x$.
A: A minimal model of an $A_\infty$-algebra $A$ is an $A_\infty$-algebra $B$ with trivial differential together with an $A_\infty$-quasi-isomorphism $B\to A$.  This gives an isomorphism $H^*A=H^*B$, and we have $H^*B=B$ since $B$ has trivial differential.  Together these show that the underlying vector space of $B$ is indeed canonically isomorphic to the homology of $A$.
(Instead of $B$, you wrote $HA$ for the minimal model.  That's of course correct in a sense, as the underlying vector space of the minimal model is indeed canonically the homology of $A$, but it is sure to cause confusion when explaining the above.)
