Can the conjugate $\phi^{-1} \circ A \circ \phi$ be linear for nonlinear $\phi$?

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

$$\phi^{-1} \circ A \circ \phi = A.$$