This question is related to this question. Let $(C^{\bullet},d)$ be a cochain complex over a field $k$ with char$k$=0. We fix an Artin local $k$-algebra $(A,m)$. A deformation of $d$ is a new operator $d+\epsilon: C^{\bullet}\otimes_k m\to C^{\bullet+1}\otimes_k m$ such that $(d+\epsilon)^2=0$. In other words we have a $\epsilon\in End^1(C^{\bullet})\otimes_k m$ such that $$ d_{End(C)}\epsilon+\frac{1}{2}[\epsilon,\epsilon]=0, $$ where $[-,-]$ is the graded commutator in $End^{\bullet}(C^{\bullet})\otimes_k m$ and $d_{End(C)}$ is the obvious differential. If we consider $(End^{\bullet}(C^{\bullet}), d_{End(C)})$ as a differential graded Lie algebra, then the above equation means that $\epsilon$ is a Maurer-Cartan element in $(End^{\bullet}(C^{\bullet})\otimes_k m, d_{End(C)})$.
For two Maurer-Cartan elements $\epsilon$ and $\eta$, we call they are gauge equivalent if there exists a $\phi\in End^0(C^{\bullet})\otimes_k m$ such that $$ e^{\phi}\circ (d+\epsilon)\circ e^{-\phi}=d+\eta. $$
Now we consider the special case that $(C^{\bullet},d)$ is acyclic, i.e. all its cohomologies vanish. Let $\epsilon$ be a deformation of such a $(C^{\bullet},d)$.
My question is: Is $\epsilon$ gauge equivalent to $0$? In other words, can we always find a $\phi\in End^0(C^{\bullet})\otimes_k m$ such that $$ e^{\phi}\circ (d+\epsilon)\circ e^{-\phi}=d? $$
