= Solution
For a <Hermitian operator> $Q$, the <operator exponential> $C_\lambda=e^{i\lambda Q}$ is a <unitary operator>, with $C_\lambda^\dagger=e^{-i\lambda Q}$. Differentiation and $[Q,C_\lambda]=0$ give
$$
\frac{d}{d\lambda}(C_\lambda A C_\lambda^\dagger)
=iQC_\lambda A C_\lambda^\dagger-iC_\lambda A C_\lambda^\dagger Q
=iC_\lambda[Q,A]C_\lambda^\dagger.
$$
When $[Q,A]=cA$, this is the first-order <ordinary differential equation> $A'(\lambda)=icA(\lambda)$ with $A(0)=A$. Multiplication by $e^{-ic\lambda}$ makes its derivative zero, proving
$$
\boxed{A(\lambda)=e^{ic\lambda}A.}
$$
Equivalently, the <commutator expansion for exponential conjugation> has successive nested <commutators> $c^r A$, whose <exponential series> sums to this expression.
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