Unipotent elements vs. unipotent linear transformations Let $n\ge 2$ and $A\in GL(n,\mathbb R)$ be a matrix and define a linear transformation on $M(n,\mathbb R)$ as follows:
$$f_A: M(n,\mathbb R) \to M(n,\mathbb R),
B \to ABA^{-1}.
$$
Suppose $f_A$ is a unipotent linear transformation on $M(n,\mathbb R) \cong \mathbb R^{n^2}$, namely the only eigenvalue of $f_A$ is $1$. Do we necessarily have that the matrix $A$'s eigenvalues are the same? (scalar multiple of a unipotent)
Conversly, if $A$ is a unipotent matrix, do we necessarily have that $f_A$ is a unipotent linear transformation?
Update:
Many thanks to the comments below about the issue of unipotency, corrected a little.
 A: I've started to write an answer to this twice now, so I'm going to make sure I finish this time.
The action of $A\in \mathrm{GL}_n(k)$ on $M_n(k)$ is given by a Kronecker product. The precise product is the Kronecker product of $A$ with its dual $A^*$, which can be thought of as $(A^{-1})^t$, where $t$ denotes the transpose. We will see that the precise form of the Kronecker product isn't so important.
The eigenvalues of the Kronecker product are simply all products of the eigenvalues of the matrices. If the eigenvalues of the product are all $1$, then the eigenvalues of each factor are all the same. You see that you don't need to know exactly what the other factor was, just that $A$ was a factor, to determine the one direction of your question.
Conversely, if $A$ is unipotent, then so is $A^*$, and so their Kronecker product is as well. Notice that even if we replace $A$ by its product with some scalar, the Kronecker product is the same, as it should be.
(There is a more conceptual answer involving Lie theory, but I guess you didn't want that since you didn't couch the question in those terms.)
