Can the conjugate $\phi^{-1} \circ A \circ \phi$ be linear for nonlinear $\phi$? More specifically, let $f：ℂ^n\to ℂ^n, f(z)=Az$ be linear and $ϕ：ℂ^n → ℂ^n$ entire and bijective, i.e. $ϕ∈\text{Aut}(ℂ^n)$. Is it possible that $ϕ^{-1}∘f∘ϕ$ is linear again when $ϕ$ is non-linear?
For example if  $ϕ(z) = Lz+c$ is affine, then
$$ ϕ^{-1}∘f∘ϕ (z) = L^{-1}ALz + L^{-1}(A-I)c $$
is linear if and only if $c \in \ker(A-I)$. Are there more sophisticated examples where $\phi$ is truly non-linear, in the sense of having non-constant derivative?
Question: Does there exist a triple $(A, B, ϕ)$, such that

*

*$A≠B$ are complex $n×n$ matrices.

*$ϕ∈\text{Aut}(ℂ^n)$ with $ϕ'$ non-constant

*$ ϕ^{-1} ∘ A ∘ ϕ=B $

*$A∘ϕ$ is nonlinear (this avoids certain trivial examples)

In particular, is there an example where $A$ and $B$ are non-similar?

EDIT 2022:

*

*Non-similarity is impossible as per Dap's comment ✔

*I added the 4th condition to rule out certain trivial examples (see comments in Qiaochu Yuan's answer)

*Dap provided an example in the comments with $A=B$

*Note that if $(A, B, ϕ)$ is a valid triple, with $B=S^{-1}AS$, then we can actually non-linearly conjugate $A$ to any matrix $C=T^{-1}AT$ similar to $A$ via $ψ = ϕ∘S^{-1}∘T$. So we immediately get an example with $A≠B$ out of Dap's contraction.

*An interesting follow-up question would be: Can we construct examples for generic $A$? So far we only have an example where $A$ is similar to a diagonal matrix and all eigenvalues have magnitude $1$.

 A: Yes. For starters, let's take $n = 2$ and
$$\phi : \left[ \begin{array}{c} z \\ w \end{array} \right] \mapsto \left[ \begin{array}{c} z \\ w + f(z) \end{array} \right]$$
where $f$ is an analytic function $f : \mathbb{C} \to \mathbb{C}$ to be chosen later. $\phi$ is analytic with inverse
$$\phi^{-1} : \left[ \begin{array}{c} z \\ w \end{array} \right] \mapsto \left[ \begin{array}{c} z \\ w - f(z) \end{array} \right].$$
Now let $A = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$ and let's write out what $\phi^{-1} \circ A \circ \phi$ is. After a little computation we get
$$\phi^{-1} \circ A \circ \phi : \left[ \begin{array}{c} z \\ w \end{array} \right] \mapsto \left[ \begin{array}{c} az + bw + b f(z) \\ cz + dw + d f(z) - f(az + bw + b f(z)) \end{array} \right].$$
To get rid of the nonlinear part of the first component we need to set $b = 0$. From here, to get rid of the nonlinear part of the second component the easiest option is to set $d = a = f(0) = 0$. So, explicitly, we can set $A = \left[ \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right]$ and $f(z) = z^2$, giving
$$\phi^{-1} \circ A \circ \phi = A.$$
