This algebra is defined on the permutation module of the symmetric group. It was studied  by K. Harada and R. Griess in the 1970s and a proof that its automorphism group is the symmetric group can also be found in Haraada's [paper][1] (see also this [one][2]) and the appendix of [this paper][3] by Dong and Griess, which explains some connections with vertex operator algebras. See also remarks in Griess's paper [here][4].

This algebra, call it $E_{n}$, has other interesting properties. Let $\tau$ be the Killing type form defined by $\tau(x, y) = \operatorname{tr} L(x)L(y)$ where $L(x)y = xy$ is the multiplication operator. Note that $\operatorname{tr} L(x) =0$ for all $x$ (I call such algebras *exact*). The symmetric bilinear form $\tau$ is nondegenerate and invariant in the sense that $\tau(xy, z) = \tau(x, yz)$ (this is somehow analogous to the properties of the Killing form of a semisimple Lie algebra). The algebra $E_{n}$ is determined up to isometric automorphism by the associated harmonic cubic polynomial $\tau(xx, x)$. This algebra is *conformally associative* in the sense that its associator $[x, y, z]= (xy)z - x(yz)$ satisfies
$$
\tau([x, y, z], w) =c\left(\tau(x, w)\tau(y,z) - \tau(x,z)\tau(y, w)\right)
$$
where $c$ is the appropriate (dimension dependent) constant that I am too lazy to calculate right now. One can prove that up to isometric automorphism there is a unique $n$-dimensional commutative nonassociative algebra that thas the properties that a. $\operatorname{tr} L(x) =0$ for all x, b. its Killing type trace form $\tau$ is nondegenerate and invariant and c. it is conformally associative. (If you contact me directly I can provide you a preprint of mine where this is proved as part of a larger project.)

Another place these algebras $E_{n}$ occur is the following. Consider the Jordan algebra of real symmetric $n\times n$ matrices. Consider its *deunitalization* that comprises trace-free symmetric matrices with the product 
$$x \star y = \tfrac{1}{2}\left(xy + yx - \tfrac{1}{n}\operatorname{tr}(xy)I\right). $$
This is a commutative nonassociative algebra such that $\operatorname{tr}L(x) =0$ for all $x$ and the Killing type trace-form $\tau(x, y) = \operatorname{tr}L(x)L(y)$ is invariant (in fact it is a constant multiple of $\operatorname{tr}(xy)$). Although this algebra is far from being conformally nonassociative in the above sense, its subalgebra of (trace-free) diagonal matrices is isomorphic to the algebra $E_{n-1}$. Moreover, every every idempotent element of the deunitalized Jordan algebra is in the orbit of an idempotent of $E_{n-1}$. This can be used, for example, to prove that the deunitalized Jordan algebras are simple. (All this is written in my aforementioned preprint; contact me if you would like details).


  [1]: https://repository.dl.itc.u-tokyo.ac.jp/?action=pages_view_main&active_action=repository_view_main_item_detail&item_id=39621&item_no=1&page_id=28&block_id=31
  [2]: https://www.sciencedirect.com/science/article/pii/0021869384901340
  [3]: https://www.sciencedirect.com/science/article/pii/S0021869398974981
  [4]: http://www.math.lsa.umich.edu/~rlg/researchandpublications/pdffiles/gnavoaI.pdf