Do distinct idempotent measures on finite binary systems have distinct supports? Suppose that $(S,*)$ is a finite set equipped with a binary operation.  Extend the binary operation to the vector space $V$ with basis $S$.
The set of probability measures on $S$, viewed as a compact convex subset of $V$ is closed under $*$
and, since $*$ is continuous, there are idempotent measures on $S$.
Must two idempotent measures on $S$ have distinct supports?
I am also interested in the more general question where the assumption of finiteness is dropped and one considers the extension (by convolution) of $*$ to the family of all finitely additive measures on $S$ (in that context, define the support of a measure to be all subsets of $S$ with positive measure).
 A: No if I understood. Take the two element left zero semigroup. All measures are idempotent. 
Added a left zero semigroup is one satisfying the identity xy=x 
Added A finite semigroup $S$ satisfies that distinct idempotent measures have distinct support iff for all idempotents $e,f\in S$ one has $SeS=SfS$ implies $e=f$.
A finite semigroup has a unique idempotent measure with full support iff it is a finite group.
Here is how the proof goes. Suppose $P$ is an idempotent measure on $S$ and assume the support of $P$ is $S$.
Claim 1: $S$ contains no ideal.
Proof.  Obviously every state is recurrent for both the right random walk an left random walk on $S$ driven by $P$ because $P$ is idempotent and the support is $S$ (and hence $P$ is stationary for these walks).  If $I$ was a proper ideal, then the states in $S\setminus I$ would have to be transient because they fall into $I$ with positive probability.  Thus $S$ has no proper ideals.
By Rees's theorem, if $e$ is an idempotent of $S$ then $eSe$ is a group $G$, there are sets $A$ and $B$, and a mapping $P\colon B\times A\to G$ such that $S\cong A\times G\times B$ with multiplication $$(a,g,b)(a',b',b') = (a,gP(b,a')g',b).$$
It is trivial to check that if $Q$ is a measure on $A$, $U$ is uniform measure on $G$ and $R$ is a measure on $B$, then $Q\times U\times R$ is an idempotent measure on $S$.  The fact that all idempotent measures $A\times G\times B$ are of this form is an easy calculation using  that the only idempotent measure on a finite group with global support is the uniform measure.    The key point is to first show that the measure is uniform on subsets of the form $\{a\}\times G\times \{b\}$ using the group result.
