= Solution
\b[No.] Let $x_0=0$ and, for $0<|x|<1$, define
$$
\phi_n(x)=|x|^n,
$$
while
$$
\psi_0(x)=1+|x|^{1/2},
\qquad
\psi_n(x)=|x|^n\quad(n\geq1).
$$
Both families are asymptotic sequences, and $\psi_n\sim\phi_n$ for every $n$.
Take $f(x)=1$. It has the exact expansion
$$
f(x)\sim1\cdot\phi_0(x)+\sum_{n=1}^{\infty}0\cdot\phi_n(x).
$$
Suppose it had an expansion $f\sim\sum_{n\geq0}b_n\psi_n$. The order-zero condition forces $b_0=1$. At order one we would then require some constant $b_1$ such that
$$
1-\psi_0-b_1\psi_1
=-|x|^{1/2}-b_1|x|=o(|x|).
$$
After division by $|x|$, the left side is $-|x|^{-1/2}-b_1$, which cannot tend to zero. This contradiction proves that no such coefficients exist. Thus <termwise equivalent asymptotic scales need not preserve expansions>.
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