Theorem 2 below offers a sufficient condition for convergence to the nontrivial equilibrium that you are referring.
The results are contingent on the uniqueness of solutions to this infinite-dimensional ODE -- I will include the proofs when possible, as needed, but, and unless I am missing something, they are quite simple.
Question. Let $\left(a_n(t)\right)_{n\in\mathbb{N}}$ be the solution to the infinite-dimensional ODE
$$a_n'(t)=a_{n+1}^2(t)-a_n(t) \,\,\,\,(\star)$$
for $n\in\mathbb{N}$, with initial condition $\left(a_n(0)\right)_{n\in\mathbb{N}}$. For which initial conditions, the solution $\left(a_n(t)\right)_{n\in\mathbb{N}}$ converges to an equilibrium distinct from the origin and the all 1's equilibrium?
Remark 1 [Domain]. The cube $\left[0,1\right]^{\mathbb{N}}$ is invariant to this dynamics. That is, if we initialize the system within this infinite-dimensional cube, then the solutions will remain there. You cannot impose condition i) as in your question. But, if you assume $a_n(0)\in\left[0,1\right]$ for all $n$, then $a_n(t)\in\left[0,1\right]$ for all $n,t$.
Remark 2 [Equilibria parametrization]. The equilibria is given by $\mathcal{E}=\left\{\left(a_{n}\right)_{n\in\mathbb{N}}\,:\,a_{n+1}^2=a_n\mbox{ for all }n\right\}$ which can be rewritten as $\mathcal{E}=\left\{\left(b^{2^{-(n-1)}}\right)_{n\in\mathbb{N}}\,:\,\mbox{ for all }b\in\left[0,1\right]\right\}$. Using your parametrization, we can further write it as $\mathcal{E}=\left\{\left(e^{-\mu 2^{-(n-1)}}\right)_{n\in\mathbb{N}}\,:\,\mbox{ for all }\mu\geq0\right\}\cup \left\{\mathbf{0}\right\}$. In other words, given any $\mu\geq 0$, then the sequence $\left(e^{-\mu 2^{-(n-1)}}\right)_{n\in\mathbb{N}}$ is an equilibrium for $(\star)$. That is, this provides a one-dimensional parametrization for the equilibria of the system. In particular, let us define for simplicity
$${\sf eq}_{\mu}\overset{\Delta}=\left(e^{-\mu 2^{-(n-1)}}\right)_{n\in\mathbb{N}}$$
the equilibrium associated with the parameter $\mu\geq 0$.
Remark 3 [Causal structure]. Observe that the evolution of $\left(a_m(t)\right)_{t\geq 0}$, for $m\geq N$, does not depend on the evolution of $\left(a_n(t)\right)_{t\geq 0}$ for any $n<N$. The state variable $a_{n+1}$ impacts $a_n$ but not the other way around. This implies the following: the tail of the initial condition is what determines the asymptotic behavior of this dynamical system. In other words, for any $N$, the sub-sequence of the initial condition $\left(a_n(0)\right)_{n\leq N}$ is irrelevant for the asymptotic behavior of the system.
The next result reveals an important property of this ODE.
Lemma 1[Monotonicity]. If $a_n(0)\leq \overline{a}_n(0)$ for all $n\in\mathbb{N}$, then $a_n(t)\leq \overline{a}_n(t)$ for all $n\in\mathbb{N}$ and $t\geq 0$.
We have an immediate corollary to Lemma 1.
Corollary 1 [Invariant sub-regions]. Let $\mu_1<\mu_2$. If there exists $N\in\mathbb{N}$ so that $e^{-\mu_2 2^{-(n-1)}}\leq a_{n}(0)\leq e^{-\mu_1 2^{-(n-1)}}$ for all $n\geq N$, then $e^{-\mu_2 2^{-(n-1)}}\leq \lim\inf_{t\rightarrow \infty}a_{n}(t)\leq \lim\sup_{t\rightarrow \infty}a_{n}(t)\leq e^{-\mu_1 2^{-(n-1)}}$ for all $n\in\mathbb{N}$.
As a consequence to Corollary 1, if the initial condition $\left(a_n(0)\right)_{n\in\mathbb{N}}$ is bounded as $e^{-\mu_2 2^{-(n-1)}}\leq a_{n}(0)\leq e^{-\mu_1 2^{-(n-1)}}$ eventually, i.e., for $n\geq N$ for some $N$, then the dynamical system $\left(a_n(t)\right)_{n\in\mathbb{N}}$ cannot converge to ${\sf eq}_{\mu}$ for any $\mu\notin \left[\mu_1,\mu_2\right]$.
Theorem 1 [Monotonicity 2]. If $a_n(0)<a_{n+1}(0)^2$ for all $n$, then $a_n(t)$ is increasing in $t$ for all $n$, i.e., $a_{n}(t)>a_{n}(t')$ for any $t>t'$ and for any $n$.
As a corollary to Theorem 1, and to the fact that $a_n(t)\in\left[0,1\right]$ for all $t$, we have convergence to an equilibrium whenever the initial condition is given by $a_n(0)<a_{n+1}(0)^2$ for all $n$. In particular, via combining Corollary 1 and Theorem 1, we have the following sufficient condition for convergence to a nontrivial equilibrium.
Theorem 2 [Sufficient condition]. If $a_n(0)<a_{n+1}(0)^2$ for all $n$, and further $e^{-\mu_2 2^{-(n-1)}}<a_n(0)<e^{-\mu_1 2^{-(n-1)}}$ eventually for some $\mu_1,\mu_2\in\left(0,\infty\right)$, then there exists $\mu\in\left[\mu_1,\mu_2\right]$ so that $a_n(t)\overset{t\rightarrow \infty}\longrightarrow e^{-\mu2^{-(n-1)}}$ for all $n$.
--------------------------- Other results --------------------------------------
Lemma 2[Invariance of the diagonal]. If $a_n(0)=b\in \left[0,1\right]$ for all $n$, then $a_n(t)=a_m(t)$ for all $n,m$. In particular, $a_n'(t)=a_n(t)^2-a_n(t)$ for all $n$.
The proof is trivial, but relies on the uniqueness of solutions to the ODE $(\star)$.
Theorem 3 [Origin's attraction]. If $\sup_{n\in\mathbb{N}} a_n(0)<1$, then, $a_n(t)\overset{t\rightarrow \infty}\longrightarrow 0$ for all $n\in\mathbb{N}$.
Proof. Consider the sequence $\left(\overline{a}_n(0)\right)_n$ so that $\overline{a}_n(0)=\sup_{n\in\mathbb{N}}a_n(0)=:c\in\left(0,1\right)$ for all $n$. Then, from Lemma 1, we have that $\overline{a}_n(t)\geq a_n(t)$ for all $n$ and $t\geq 0$. From Lemma 2, we have that $\overline{a}'_n(t)=\overline{a}_n(t)^2-\overline{a}_n(t)$. Therefore, $\overline{a}_n(t)\overset{t\rightarrow \infty}\longrightarrow 0$ for all $n$. Thus, $a_n(t)\overset{t\rightarrow \infty}\longrightarrow 0$ from the boundedness and since $\left[0,1\right]^{\mathbb{N}}$ is invariant to the infinite-dimensional dynamical system $(\star)$.
Theorem 4 [1's attraction]. If $a_n(0)=1$, for all $n>N$ for some $N\in\mathbb{N}$ large enough, then $a_n(t)\overset{t\rightarrow \infty}\longrightarrow 1$ for all $n\in\mathbb{N}$.
Proof. It is trivial to check that $a_n(t)=1$ for all $t\geq 0$ for all $n>N$. we have $a_n'(t)=1-a_n(t)$ for $n=N$ and hence $a_n(t)\overset{t\rightarrow \infty}\longrightarrow 1$. From here, it is trivial to check that $a_n(t)\longrightarrow 1$ for all $n$.
