I am looking for a square $n \times n$ matrix $A$ such that its spectral moments follow a geometric progression: $Tr(A^r) = k^r$, for some $k$. I believe this to be impossible since $Tr(A^0) = Tr(I) = n$. So perhaps I should be looking for something like $$ Tr(A^r) = n^{r+1}, $$ or something like $$ Tr(A^r) = nk^{r}. $$
I'm interested in complex or real matrices, though other fields would be acceptable. Apologies if the question sounds vague, but I am dealing with an open ended question here!
What I have tried: I know that the coefficients of the characteristic polynomial can be expressed in terms of the spectral moments using Newton's identities. In this way, I can fix the traces of $A,\ldots,A^n$ to be whatever I want, find the corresponding polynomial, and find a matrix whose characteristic polynomial matches it. However, this does not give me any control on the spectral moments of powers greater than $n$. So I'm beginning to wonder if this is at all possible.
Also note that for $n=1$ this is trivial since $A$ is just an element of the underlying field and we have $Tr(A^r) = A^r$. I'm interested in $n>1$.
EDIT: if possible, I would also like $A$ to be invertible...
Claim Over characteristic-0 field $\mathbb{F}$ (such as $\mathbb{R},\mathbb{C}$), if $A$ satisfies the condition, then $A$ has only one eigenvalue $\lambda$ in $\bar{\mathbb{F}}$.
Proof: Suppose $\operatorname{Tr}A^k=nr^k$.
If $r\neq 0$, we find $\operatorname{Tr}(r^{-1}A)^k=n$ for all $k$. We calculate the coefficient $c_k$ in the characteristic polynomial $\chi_{r^{-1}A}(t)=\det(tI-r^{-1}A)=\sum_{k=0}^n c_k t^k$ by \begin{align*} c_{n-k}&=\frac{(-1)^k}{k!}\begin{vmatrix} \operatorname{Tr}(r^{-1}A) & k-1\\ \operatorname{Tr}(r^{-1}A)^2 & \operatorname{Tr}(r^{-1}A) & k-2\\ \vdots & \vdots & \vdots & \ddots \\ \operatorname{Tr}(r^{-1}A)^{k-1} & \operatorname{Tr}(r^{-1}A)^{k-2} &\dots & \dots & 1\\ \operatorname{Tr}(r^{-1}A)^k & \operatorname{Tr}(r^{-1}A)^{k-1} &\dots & \dots & \operatorname{Tr}(r^{-1}A)\\ \end{vmatrix}\\ &= \frac{(-1)^k}{k!}\begin{vmatrix} n & k-1\\ n & n & k-2\\ \vdots & \vdots & \vdots & \ddots \\ n & n &\dots & \dots & 1\\ n & n &\dots & \dots & n\\ \end{vmatrix}\\ &= \frac{(-1)^k}{k!}\begin{vmatrix} (n-k+1) & k-1\\ 0 & n & k-2\\ \vdots & \vdots & \vdots & \ddots \\ 0 & n &\dots & \dots & 1\\ 0 & n &\dots & \dots & n\\ \end{vmatrix}\\ &=\frac{(n-k+1)}{k}(-1)c_{n-k+1} \end{align*} So starting from $c_n:=1$, we have $c_{n-1}=-n, c_{n-2}=\binom{n}2, c_{n-3}=-\binom{n}3, \dots$ and so the characteristic polynomial of $r^{-1}A$ is $\chi_{r^{-1}A}(t)=(t-1)^n$. Hence $\chi_A(t)=(t-r)^n$.
If $r=0$, then all elementary polynomials in the eigenvalues are $0$ from Newton's identity, so $\chi_A(t)=t^n$. QED.
So the only possibilities of $A$, when working in characteristic 0, are of the form $$ A=rI+N $$ where $N$ is nilpotent.
Over positive characteristic you have other possibilities. For example, over $\mathbb{F}_p$ and $n=2p$, $p>2$, any matrix of the form $$ A=\begin{pmatrix} aI+U & C\\ 0 & bI+V \end{pmatrix} $$ for any $a,b\in\mathbb{F}$, any strictly upper-triangular $U,V$ and any $C$ will have $\operatorname{Tr}(A^k)=0$ for $k>0$.