$(\mbox{congruence class of }A) \cap (\mbox{conjugacy class of }A)=\{A\},$ possible for non-scalar matrix $A$? We consider matrices below in the ring of square matrices $M_n(K)$, over field $K$.


*

*$A,B$ are said to be congruent if $P^tAP=B$ for some $P\in M_n(K)$.

*$A,B$ are said to be conjugate if $P^{-1}AP=B$ for some invertible $P\in M_n(K)$.
As usual, by congruent class and conjugacy class of a matrix $A$, we mean its equivalence class under corresponding equivalence relation.
Of course, these two classes may be very different; but how long? I will put the following question for this:

If $A$ is a non-scalar matrix in $M_n(K)$ then is it possible that
  $$(\mbox{congruence class of }A) \cap (\mbox{conjugacy class of }A)=\{A\}?$$

 A: Edit. When the characteristic of the field is not $2$, the intersection cannot be a singleton.
Since $A$ is not a scalar matrix, its size is at least $2\times2$. Now suppose the field's characteristic is not $2$. There are two possibilities:


*

*If $A=aI+bE$, where $a,b$ are scalars, $E$ is the all-one matrix and $b\ne0$ (by assumption, $A$ is not a scalar matrix), then $P^{-1}AP=P^TAP\ne A$ when $P=\operatorname{diag}(-1,1,\ldots,1)$. Therefore the intersection is not a singleton.

*If $A$ is not in the form of $aI+bE$, then $P^{-1}AP=P^TAP\ne A$ for some permutation matrix $P$. Hence the intersection of the two equivalence classes is not a singleton.


What if the field has characteristic $2$? For instance, if the field is $GF(2)$, then from case 2 in the above, we see that in order that the intersection is a singleton, $A$ must be equal to $E$ or $I+E$. Now, one can verify by brute force that the intersection is indeed a singleton when $n=2$ and $A=I+E$ (in fact, in this case, it happens that $P^T(I+E)P=I+E$ for every invertible matrix $P$), but I don't know if singleton examples exist for any larger $n$.
