How do I get f' from f that are A-equivalent?

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I'm struggling with following problem realted to the singularity theory. I assume that $f, f' \in C^\infty(X,Y)$ and $f$ is a stable mapping. I would like to move from $f$ to some $f'$ which is $A$-equivalent to $f$. So there should exist $\psi\in \text{Diff}(Y)$ and $\varphi \in \text{Diff}(X)$ such that $f' = \psi \circ f \circ \varphi^{-1}$. The idea to compute them is to take one-parameter families of diffeomorphisms such that $f_t = \psi_t\circ f \circ \varphi_t^{-1}$. Differentiating this with respect to time gives me an ODE $$ \frac{df_t(x)}{dt} = df_t(x)v(x) + w(f_t(x)),$$ with $f_0(x) = f(x), f_1(x) = f'(x)$. What I need to do is to design the vector fields $v$ and $w$. Is my reasoning correct? It seems to me that integrating this ODE is quite hard. Is there a different way of solving this problem?

I tried to find the solution in the books of Golubitsky and Guillemin, Arnold and Gusein-Zade and Varchenko but I could not figure out how to solve my problem. Could you point me to the right direction?