I was scratching my head over the following for almost an hour today, and since I don't have a blog to share the resolution on, I'll post it here.
Let $\mathbb{T} = (T,\eta,\mu)$ be a monad on a category $\mathcal{C}$. Consider the following instances of the unit identities at objects $M$ and $T(M)$ of $\mathcal{C}$: $$\require{AMScd} \begin{CD} T(M) @>{\eta_{TM}}>> T^2(M) @<{T\eta_M}<< T(M) \\ @| @VV{\mu_M}V @|\\ T(M) @= T(M) @=T(M) \end{CD} \text{ } \begin{CD} T^2(M) @>{T\eta_{T(M)}}>> T^3(M)\\ @| @VV{\mu_{T(M)}}V\\ T^2(M) @= T^2(M) \end{CD}$$ By naturality of $\mu$ we also have the following square: $$\begin{CD} T^2(M) @>{T\eta_{T(M)}}>> T^3(M)\\ @VV{\mu_{M}}V @VV{\mu_{T(M)}}V\\ T(M) @>{\eta_{T(M)}}>> T^2(M). \end{CD}$$ From this we conclude that $\eta_{T(M)}$ is both a left and right adjoint to $\mu_M$, and so (since $M$ was arbitrary), $\mu$ is a natural isomorphism and hence $\mathbb{T}$ is an idempotent monad...
Obviously not all monads are idempotent (your favourite monad probably isn't for obvious reasons). So what's wrong here?
Never forget that naturality diagrams only apply to morphisms in the images of the functors involved! For a monad, the multiplication is a natural transformation $\mu:T^2 \to T$, so since $\eta_{T(M)}$ typically isn't of the form $T(g)$ for some morphism $g$ in $\mathcal{C}$, there is no requirement for the lower square to be commutative in general. In fact, it is not much of a stretch to realise that commutativity of that square for every object $M$ is equivalent to idempotency of the monad.