The Willmore energy $\mathcal{W} = \int_M H^2 dA$ differs from the functional $$\widetilde{\mathcal{W}} = \int_M (H^2-K) dA$$ just by a constant as one can see fron the Gauss - Bonnet theorem ($K$ here is the Gaussian curvature of $M$).

The expression $H^2-K$ in $\widetilde{\mathcal{W}}$ is the half of the square of the length of the trace-free part of the second fundamental form which is a (pointwise) conformally invariant density of conformal weight $-2$, while "dA" can be seen as a density with conformal weight $2$, so the entire integrand $(H^2-K) dA$ is independent of a choice of a metric.

Thus $\widetilde{\mathcal{W}}$ is manifestly conformally, and in particular, Möbius invariant. So is $\mathcal{W}$.

(A Liouville's theorem ensures that conformal maps of $\mathbb{R}^n$, $n\ge 3$, are restrictions of Möbius transformations.)

Edit. The above is an attempt to address the original request for a "tweetable" argument.

Of course, the precise statement is that the Willmore energy is conformally invariant with respect to the conformal transformations of the ambient space. (Otherwise we would not be able to invoke the Liouville's theorem).

The correct definition of the Willmore energy involves an immersion $f\colon M \rightarrow \mathbb{R}^3$ and the induced conformal structure on the immersed manifold. The Dirac spheres show, in particular, that the Willmore energy does depend on the immersion.

The Willmore energy $\mathcal{W} = \int_M H^2 dA$ differs from the functional $$\widetilde{\mathcal{W}} = \int_M (H^2-K) dA$$ just by a constant as one can see from the Gauss - Bonnet theorem ($K$ here is the Gaussian curvature of $M$).

The expression $H^2-K$ in $\widetilde{\mathcal{W}}$ is the half of the square of the length of the trace-free part of the second fundamental form which is a (pointwise) conformally invariant density of conformal weight $-2$, while "dA" can be seen as a density with conformal weight $2$, so the entire integrand $(H^2-K) dA$ is independent of a choice of a metric.

Thus $\widetilde{\mathcal{W}}$ is manifestly conformally, and in particular, Möbius invariant. So is $\mathcal{W}$.

(A Liouville's theorem ensures that conformal maps of $\mathbb{R}^n$, $n\ge 3$, are restrictions of Möbius transformations.)

Edit. The above is an attempt to address the original request for a "tweetable" argument.

Of course, the precise statement is that the Willmore energy is conformally invariant with respect to the conformal transformations of the ambient space. (Otherwise we would not be able to invoke the Liouville's theorem).

The correct definition of the Willmore energy involves an immersion $f\colon M \rightarrow \mathbb{R}^3$ and the induced conformal structure on the immersed manifold. The Dirac spheres show, in particular, that the Willmore energy does depend on the immersion.

The Willmore energy $\mathcal{W} = \int_M H^2 dA$ differs from the functional $$\widetilde{\mathcal{W}} = \int_M (H^2-K) dA$$ just by a constant as one can see fron the Gauss - Bonnet theorem ($K$ here is the Gaussian curvature of $M$).

The expression $H^2-K$ in $\widetilde{\mathcal{W}}$ is the half of the square of the length of the trace-free part of the second fundamental form which is a (pointwise) conformal invariant density of conformal weight $-2$, while "dA" can be seen as a density with conformal weight $2$, so the entire integrand $(H^2-K) dA$ is independent of a choice of a metric.

Thus $\widetilde{\mathcal{W}}$ is manifestly conformally, and in particular, Möbius invariant. So is $\mathcal{W}$.

(A Liouville's theorem ensures that conformal maps of $\mathbb{R}^n$ are restrictions of Möbius transformations.)

The Willmore energy $\mathcal{W} = \int_M H^2 dA$ differs from the functional $$\widetilde{\mathcal{W}} = \int_M (H^2-K) dA$$ just by a constant as one can see fron the Gauss - Bonnet theorem ($K$ here is the Gaussian curvature of $M$).

The expression $H^2-K$ in $\widetilde{\mathcal{W}}$ is the half of the square of the length of the trace-free part of the second fundamental form which is a (pointwise) conformally invariant density of conformal weight $-2$, while "dA" can be seen as a density with conformal weight $2$, so the entire integrand $(H^2-K) dA$ is independent of a choice of a metric.

Thus $\widetilde{\mathcal{W}}$ is manifestly conformally, and in particular, Möbius invariant. So is $\mathcal{W}$.

(A Liouville's theorem ensures that conformal maps of $\mathbb{R}^n$, $n\ge 3$, are restrictions of Möbius transformations.)

Edit. The above is an attempt to address the original request for a "tweetable" argument.

Of course, the precise statement is that the Willmore energy is conformally invariant with respect to the conformal transformations of the ambient space. (Otherwise we would not be able to invoke the Liouville's theorem).

The correct definition of the Willmore energy involves an immersion $f\colon M \rightarrow \mathbb{R}^3$ and the induced conformal structure on the immersed manifold. The Dirac spheres show, in particular, that the Willmore energy does depend on the immersion.