= Solution
A one-dimensional commutative <formal group law> over a ring $R$ is a series $F(X,Y)\in R[[X,Y]]$ satisfying
$$
F(X,0)=X,
\qquad F(X,Y)=F(Y,X),
\qquad F(F(X,Y),Z)=F(X,F(Y,Z)).
$$
An isomorphism $u:F\to G$ is a series $u(T)\in TR[[T]]$ with a compositional inverse and
$$
u(F(X,Y))=G(u(X),u(Y)).
$$
The multiplication series is defined recursively by $[0]_F=0$, $[1]_F=T$ and $[n+1]_F=F([n]_F,T)$, with the <formal inverse> handling negative $n$. Its linear term is
$$
[n]_F(T)=nT+O(T^2).
$$
By the <invertible morphism criterion for formal group laws>, it is an isomorphism exactly when its linear coefficient $n$ is a unit of $R$. Indeed, when $n\in R^\times$, recursive coefficient comparison constructs a unique compositional inverse; applying the morphism identity for $[n]_F$ shows that the inverse also respects $F$. Conversely, an invertible series must have a unit linear coefficient. Therefore
$$
\boxed{[n]_F\text{ is an isomorphism}\quad\Longleftrightarrow\quad n\in R^\times.}
$$
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