Tweetable way to see Riemannian isometries are harmonic?

Not exactly 'tweetable', but perhaps the identity (1) may help, if all you want to do is avoid the Euler-Lagrange equations. For simplicity, assume that $M^n$ is oriented. (One can write the identity (1) below as an identity on densities, so the orientability hypothesis is not essential, but I'll leave that detail for the interested.)

If $g$ is a metric on $M$, let $\mathrm{vol}_g>0$ denote its associated volume form (positively oriented). Then for any vector field $X$ on $M$, one has the $n$-form identity $$ \tfrac12\,\mathrm{tr}_g\bigl(\mathcal{L}_Xg\bigr)\,\mathrm{vol}_g = \mathrm{d}\bigl(\iota(X)\,\mathrm{vol}_g\bigr),\tag1 $$ where $\mathcal{L}_X$ denotes Lie derivative with respect to $X$ and $\iota(X)\,\mathrm{vol}_g$ denotes the $(n{-}1)$-form that is the interior product of $X$ with $\mathrm{vol}_g$.

Since, for any vector field $X$ on $M$ (say, with compact support), one has, by definition $$ \left.\frac{d\ }{dt}\right|_{t=0}\bigl(\mathrm{exp}_{tX}^*g\bigr) = \mathcal{L}_Xg\,, $$ one has, for the $1$-parameter family of diffeomorphisms $\mathrm{exp}_{tX}:M\to M$, $$ \left.\frac{d\ }{dt}\right|_{t=0}\bigl(e(\mathrm{exp}_{tX})\bigr) =\left.\frac{d\ }{dt}\right|_{t=0}\bigl(\tfrac12\mathrm{tr}_g(\mathrm{exp}_{tX}^*g)\,\mathrm{vol}_g\bigr) = \tfrac12\mathrm{tr}_g(\mathcal{L}_Xg)\,\mathrm{vol}_g = \mathrm{d}\bigl(\iota(X)\,\mathrm{vol}_g\bigr). $$ Integrating this over $M$, $$ \left.\frac{d\ }{dt}\right|_{t=0}E(\mathrm{exp}_{tX}) = \int_M \left(\left.\frac{d\ }{dt}\right|_{t=0}e(\mathrm{exp}_{tX})\right) = \int_M \mathrm{d}\bigl(\iota(X)\,\mathrm{vol}_g\bigr) = 0, $$ by Stokes' Theorem. Thus, the identity $(= \mathrm{exp}_{0X})$ is $E$-critical for all compactly supported variations, as desired.

Let $u: M\rightarrow N$ be a diffeomorphism. By staring at the Dirichlet energy formula and knowing that integration by parts works just as well here as for the classical case for functions, you know that a map is harmonic if and only if $\Delta u = 0$.

However, if $u$ is isometric, then you can see in your head that $$ 0 = \nabla_k(\partial_iu\cdot\partial_ju) = \nabla^2u_{ik}\cdot\partial_ju + \partial_iu\cdot\nabla^2_{kj}u. $$ But you also know, given the symmetries of the indices that the standard argument proving the uniqueness of the Levi-Civita connection (also known as Cartan's lemma) implies that $$ \partial_ku\cdot\nabla^2_{ij}u = 0. $$ Specifically, \begin{align*} \partial_ku\cdot\nabla^2_{ij}u & = \frac{1}{2}(\nabla_i(\partial_ku\cdot\partial_ju) + \nabla_j(\partial_ku\cdot\partial_iu) - \nabla_k(\partial_iu\cdot\partial_ju))\\ & = \frac{1}{2}(\nabla_ig_{kj} + \nabla_jg_{ki} - \nabla_kg_{ij})\\ &= 0. \end{align*} Since $u$ is a diffeomophism, $\partial_1 u, \cdots, \partial_n u$ span $T_*N$ and therefore, $\nabla^2u = 0$ and $u$ is harmonic.

If $\dim M < \dim N$, then the argument above shows that $\nabla^2u$ is normal to $T_*M$. In fact, it is the second fundamental form (viewed as a normal-vector-valued symmetric tensor).