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Let $M$ be the monoid of order-preserving functions $\mathbb{N}\to \mathbb{N}$, with $\mathbb{N}=\{0,1,\dots \}$, regarded as a one-object category. The endofunctor $T$ fixes that object and sends $f$ to $Tf(0)=0$, $Tf(n+1)=f(n)+1$. The displayed maps give its monad unit and multiplication. The multiplication is not injective, so this is not an idempotent monad. Its only algebra for a monad is $\mu$, because $a\eta =1$ forces $a(n+1)=n$ and monotonicity forces $a(0)=0$. Algebra endomorphisms are exactly the functions fixing zero. The Kleisli comparison functor sends $f$ to the function which is zero at zero and equals $f(n)$ at $n+1$; its inverse sends $h$ to $n\mapsto h(n+1)$. Thus the comparison is an isomorphism of categories even though the monad is not idempotent.

A transformation monoid is a collection of self-maps of a fixed set $X$ that contains the identity function and is closed under function composition. The associative property comes from function composition. A monoid action describes transformations of a set that obey a monoid's multiplication and identity laws.