(True/false)if $f$ and $g$ are differentiable show that $g '(0)$ and $f' (0)$ have the same eigenvalues

Question

Let $f,g:U\subseteq \mathbb{R}^{m}\to \mathbb{R}^{m}$ differentiables in the open $U$ that contains $0$ and $f(0)=g(0)=0$. If $h\circ f=g\circ h$ and $h'(0): \mathbb{R}^{m}\to \mathbb{R}^{m}$ is an isomorphism. is it true that $f '(0)$ and $g' (0)$ have the same eigenvalues?

I've been trying to find a counterexample in $\mathbb{R}$ without success. Is there a counterexample?

Answer 1

By the Chain Rule, $$h'(f(0)) \circ f'(0) = (h \circ f)'(0) = (g \circ h)'(0) = g'(h(0)) \circ h'(0),$$ which is the same as saying that $h'(0) \circ f'(0) \circ [h'(0)]^{-1} = g'(h(0))$. So, if $h(0) = 0$, then $f'(0)$ and $g'(0)$ has the same eigenvalues.