Do sets of commuting elements having conjugates in a commutative submonoid have a single conjugating element? It is known that any set of commuting diagonalizable matrices is simultaneously diagonalizable. So, it would be nice to ask the following generalization:
Given a monoid $M$ (not necessarily a group), a commutative submonoid $N$ of $M$ (with the same identity, so $eMe$ for some idempotent $1 \ne e \in M$ does not count even if it happens to be commutative), and some subset $A$ of $M$ for which $\forall a,b \in A\, ab=ba$ and $\forall a \in A\, \exists p \in M^{\times}\, p^{-1}ap \in N$ (where $M^{\times}$ is the subgroup of invertible elements of $M$), is it in fact true that $\exists p \in M^{\times}\, \forall a \in A\, p^{-1}ap \in N$?
If not, is it true under some additional assumptions (e.g. $M$ is a group, $N$ is an abelian subgroup of $M$, or $A$ is finite)?
 A: None of the conditions you give suffice; that's not going to work in general. Here is a counterexample with $M$ a group, $N$ an abelian subgroup, and $A$ finite.
let $M=S_6$, let $N=\{e,(12)\}$, let $A=\{(34), (56)\}$. Note the elements of $A$ commute with each other, and they can each be conjugated into $N$ (any transposition is conjugate to $(12)$). But the elements of $A$ cannot be simultaneously conjugated into $N$, because they only have one place to go, and so that would require $g(34)g^{-1}=g(56)g^{-1}$, which is of course impossible since conjugation induces a bijection.
A: This is not true even for matrices if you extend to the infinite-dimensional case.  That is, an infinite-dimensional vector space can have a collection of diagonalizable endomorphisms which commute but are not simultaneously diagonalizable.  For instance, let $k$ be the base field, let $S$ be any infinite set, and let $V=k^S$.  For each $T\subseteq S$, let $p_T:V\to V$ be the projection onto $k^T$.  Then each $p_T$ is diagonalizable (pick a basis for $k^S$ and a basis for $k^{T\setminus S}$) and they commute ($p_Tp_{T'}=p_{T\cap T'}$).  However, it is easy to see that if $f\in V$ is an eigenvector of every $p_T$ then there is at most one $s\in S$ such that $f(s)\neq 0$.  So, the common eigenvectors of all the $p_T$s span only $k^{\oplus S}$, which is not all of $V$.
(So, in the notation of your question, this would be a counterexample with $M=\operatorname{End}(V)$, $N$ the submonoid of endomorphisms that are diagonal with respect to some fixed basis, and $A=\{p_T:T\subseteq S\}$.  If you want, you can have $M$ be a group by taking $M=GL(V)$, $N$ the subgroup of diagonal automorphisms, and $A=\{1+p_T:T\subseteq S\}$, as long as $\operatorname{char}(k)\neq 2$.)
