X,AX have no common eigenvalues
Let $A\in M_n(\mathbb{C})$ s.t. $A\not= I_n$. Show that there is $X\in M_n(\mathbb{C})$ s.t. $X$ and $AX$ have no common eigenvalues.
Perhaps it is easy to prove, but I don't know how to do.
Comment. The proof is easy when $1\notin spectrum(A)$. Moreover, if the asked result is true, then we can show that $\{X;X,AX \;\;\text{have no common eigenvalues}\}$ is Zariski open dense in $M_n(\mathbb{C})$.
This is ugly but I hope it works. Assume $A$ is already in Jordan normal form. We consider three possibilities:
(a) $A$ has not any trivial ($1\times1$) Jordan block for the eigenvalue $1$. Let $A=J\oplus F$, where $J$ is a direct sum of all nontrivial Jordan blocks for the eigenvalue $1$ and $F$ contains only Jordan blocks for other eigenvalues. Using a similarity transform, we can make the superdiagonal entries of $J$ arbitrarily large, while leaving its diagonal entries unchanged. So, we may assume that all singular values of this modified $J$ are different from $1$. Take $X=J^T \oplus I$, so that $AX=(JJ^T)\oplus F$. The only eigenvalue of $X$ is $1$, but all eigenvalues of $AX$ are different from $1$.
(b) $A$ has at least one trivial Jordan block for the eigenvalue $1$, and $A$ is not diagonalizable. Let $A=I_k\oplus J\oplus A'$, where $J$ is some $m\times m$ nontrivial Jordan block (its eigenvalue is irrelevant) and $A'$ has not any trivial Jordan block for the eigenvalue $1$. By (a), there is some $X'$, whose only eigenvalue is $1$, such that $X'$ and $A'X'$ have non-overlapping spectra. So, we only need to cope with the submatrix $I_k\oplus J$ of $A$.
When the superdiagonal entries of $J$ are sufficiently large, all singular values of $J$ are different from $1$. Therefore, by a further change of basis, we may assume that $A=I_k\oplus SV^\ast\oplus A'$, where $SV^\ast$ is a SVD with all singular values $s_1,s_2,\ldots,s_m\ne1$. Let $v$ be the first column of $V$ and let $W$ contains the other columns. Take $$ X=\pmatrix{0&I_k&0\\ v&0&W}\oplus (tX') $$ for some $t>0$. (Here the first summand is $(k+m)\times(k+m)$; the zero submatrices within it are not necessarily square. The second summand $tX'$ is $(n-k-m)\times(n-k-m)$; it has the same size as $A'$.) Since the first summand is unitary, the eigenvalues of $X$ either have unit moduli or are equal to $t$. In contrast, the eigenvalues of $$ AX=\left(\begin{array}{cc|c} 0&I_k\\ s_1&0\\ \hline &&\operatorname{diag}(s_2,\ldots,s_m) \end{array}\right)\oplus (tA'X') $$ are $s_2,\ldots,s_m$, the $(k+1)$-th roots of $s_1$, and those eigenvalues of $tA'X'$. So, when $t$ is sufficiently large, the spectra of $X$ and $AX$ do not overlap.
(c) $A$ is diagonal. Let $A=(I_k\oplus\lambda_1)\oplus\operatorname{diag}(\lambda_2,\ldots,\lambda_{n-k})$, where each $\lambda_j\ne1$. In this case we can take $X=\pmatrix{0&1\\ &\ddots&1\\ &&\ddots&\ddots\\ &&&0&1\\ c&0&\cdots&\cdots&0}\oplus I_{n-k-1}$ for some number $c\in\mathbb C$. The spectrum of $X$ contains $1$ and the $(k+1)$-th roots of $c$, while the spectrum of $AX$ contains $\lambda_2,\ldots,\lambda_{n-k}$ and the $(k+1)$-th roots of $c\lambda_1$. As $\lambda_1\ne1$, we can pick $c$ such that the two spectra do no overlap.