Adjoint action on idempotents in Clifford algebras Given a Clifford algebra ${\operatorname{Cl}_{p,q}}$, we can define the group of units of the algebra ${\operatorname{Cl}_{p,q}}^{*}$, and the adjoint action of this group on the Clifford algebra as usual: $$\operatorname{Ad}(S)(x)=SxS^{-1},$$ where $S\in{\operatorname{Cl}_{p,q}}^{*}$ and $x\in{\operatorname{Cl}_{p,q}}$.
I am working with idempotents on Clifford algebras, so I thought particularly about this action on the set of idempotents. It can be easily seen that if $f$ is an idempotent, then $SfS^{-1}$ is also an idempotent. It is also clear that if $f$ and $g$ are orthogonal, then $SfS^{-1}$ and $SgS^{-1}$ are also orthogonal, and that if $f$ is primitive then $SfS^{-1}$ is primitive.
So, I was thinking: what are the orbits of this action in the set of idempotents?, is this action transitive on primitive idempotents?
I am asking this question in the context of Clifford algebras (which are matrix algebras or direct sums of 2 copies of them), so perhaps I should look for results in matrix algebras (at least in some cases), but I will be glad to hear about results in general rings (perhaps with the adjoint action of the group of units).
PS: This question led me to another question, that is, does the set of idempotents in a general ring have any particular structure? All I found is the known poset structure.
Thanks in advance!
 A: Suppose the Clifford algebra is isomorphic to $M_n(F)$.  
It's a well-known fact of linear algebra that idempotent matrices can all be diagonalized to a matrix with $0$'s and $1$'s on the diagonal.  The primitive ones are clearly the ones which have exactly one $1$ on the diagonal.  Furthermore, any matrix of that form (zero except for a single $1$ on the diagonal) is similar to the matrix with a $1$ in the top left corner and zeros elsewhere (you just use a permutation matrix to conjugate.)
This shows that the conjugation action of $GL(n,F)$ on $M_n(F)$ acts transitively on primitive idempotents, and a yes to your question in that case.
However, the answer in the case when it is isomorphic to $M_n(F)\times M_n(F)$ is no.  The reason is that the primitive idempotents are now separated into two groups: ones which are nonzero in the left, and ones which are nonzero in the right.  Clearly it is impossible to conjugate and move between the two.
For a second way to see this, consider that the splitting into two rings gives a central idempotent $e$ that annihilates one and acts like the identity on the other.  If you could conjugate $x$ in one half to $y$ in the other half with $z$ (let's say $zxz^{-1}=y$) and $e$ annihilates the half $x$ is in, then $y=ey=ezxz^{-1}=zexz^{-1}=0$, a contradiction.

In my experience, the poset structure on the idempotents is the most common one.   There is also the Boolean algebra of central idempotents which is used in ring theory.
I believe there might be some graphs of idempotents that might be distinct from the one you get from the poset structure, but I am not very familiar with them.
