This is a repost of https://math.stackexchange.com/questions/4692364/automorphisms-of-matrix-algebras-and-picard-group (asked on MSE).
Notation. In what follows, $R$ is a commutative ring with $1$, $n\geq 1$ is an integer, $\mathcal{B}=(e_1,\ldots,e_n)$ is a fixed basis of $R^n$, and $e_{ij}$ is the usual elementary endomorphism, namely the unique endomorphism of $R^n$ such that $e_{ij}(e_k)=\delta_{j,k}e_i$.
In other words, the matrix of $e_{ij}$ wrt to $\mathcal{B}$ is the usual elementary matrix $E_{ij}$.
Context. The relationship between the outer automorphism group of the $R$-algebra ${\rm End}(R^n)$ and the $n$-torsion of the Picard group of $R$ is well-known (this can even be generalized to Azumaya algebras). The standard result is that there is a group isomorphism between ${\rm Out}({\rm End}(R^n))$ and the subgroup $\mathcal{P}_n(R)$ of ${\rm Pic}(R)$ consisting of isomorphisms classes of $R$-modules $M$ satisfying $M^n\simeq_R R^n.$ (For those who are not familiar with Picard groups: this is the group of isomorphism classes of projective modules of rank one, and the group law is induced by the tensor product). It can be shown that $\mathcal{P}_n(R)\subset {\rm Pic}_n(R)$.
In the paper Automorphisms of matrix algebras of commutative rings, Isaacs provides a direct proof of thie group isomorphism above.
The isomorphism is the usual one (the one used also in the more general case of Azumaya algebras): it sends the conjugacy class of $\theta$ to the isomorphism class of the $R$-module $S_\theta=\lbrace v\in {\rm End}(R^n)\mid u\circ v =v \circ \theta(u) \ \mbox{ for all }u\in {\rm End}(R^n) \rbrace$.
Motivation. Out of curiosity, I tried (really hard) to write down the inverse isomorphism, but I found it very difficult for various reasons. In particular, the proof of surjectivity of the map above seems really too complicated for such a little result (it uses a lot of cumbersome ad hoc computations and various intermediate results), so I have decided to come with a direct proof of my own.
My work so far.
If $\theta$ is an automorphism of $R$-algebras of $\mathrm{End}(R^n)$, set $M_\theta={\rm Im}(\theta(e_{11})).$ I can show that the isomorphism class of $M_\theta$ only depends of the conjugacy class of $\theta$.
If $M$ is an $R$-module satisfying $M^n\simeq R^n$, fix an isomorphism $\varphi: M^n\to R^n$.
Set $\iota_i:M\to M^n, x\mapsto (0,\ldots,0,x,0,\ldots,0)$, where $x$ is in position $i$, and let $\pi_j:M^n\to M$ the projection on the $j-th$ component.
Then, one may show that there is a unique $R$-algebra automorphism $\theta_M$ of ${\rm End}(R^n)$ such that $\theta_M(e_{ij})=\varphi\circ \iota_i\circ\pi_j\circ \varphi^{-1}$ for all $i,j$.
I can show that the conjugacy class of $\theta_M$ does not depend on the choice of $\varphi$ and only depends on the isomorphism class of $M$.
Hence, we get a map $\alpha: \mathcal{P}_n(R)\to {\rm Out}({\rm End}(R^n))$, sending the isomorphism class of $M$ to the conjugacy class of $\theta_M$.
I can show that $\alpha$ is a bijective map, where the inverse map $\beta$ sends the conjugacy class of $\theta$ to the isomorphism class of $M_\theta.$
Finally, here comes my question.
Question. I would like to prove that $\alpha$ is a group morphism. However, despite all my efforts I did not succeed at the moment.
I would appreciate some help on this.
Remark. I am aware that all of my work is probably Isaac's proof in disguise, but curiously, my point of view allows me to prove really easily the surjectivity of $\alpha$, that I couldn't see with Isaac's arguments.
Side Question. At the moment, the relationship between $M_\theta$ and $S_\theta$ is not clear to me. Does anyone see what it could be ?
