Outer and Inner automorphism. Let $A$ be an algebra. For $a, b \in A$ we define maps:
$L_{a}(x) = a x $ and $ R_{b} (x) = xb$
$M(A) = \{\sum^{n}_{i=1} L_{a_{i}}R_{b_{i}} : a_i , b_i \in A ,  n \in \mathbb{N}  \}$

Lemma: If $A$ is a finite dimensional central simple algebra, then $M(A) =\operatorname{End}_F (A)$.
Definition: An automorphism $ \varphi$ of a unital ring $R$ is said to be an inner automorphism if there exists an invertible $a \in R$ such that $\varphi(x) = axa^{−1}$, $x \in R$. An automorphism that is not inner is called outer.

I read two examples in connection with inner automorphism. I have a problem with each of them.
Example 1: The conjugation $z \mapsto \overline{z} $ is an automorphism of the $\mathbb{R}$-algebra $\mathbb{C}$. Of course it is outer for the identity map is obviously the only inner automorphism of a commutative ring. We also remark that it is an element of $\operatorname{End}_{\mathbb{R}}(\mathbb{C})$, but not of $M(\mathbb{C})$.

Why can we say  that  “it is an element of $\operatorname{End}_{\mathbb{R}}(\mathbb{C})$, but not of $M(\mathbb{C})$”?

Example 2: Let $S$ be a ring and let $R = S \times S$ be the direct product of two copies of $S$. Then $(s, t) \mapsto (t, s)$ is an outer automorphism of $R$.

Why is $(s, t) \mapsto (t, s)$  an outer automorphism of $R$? 

 A: For the second question, suppose $(s,t)\mapsto(t,s)$ is inner; then there exists $(a,b)\in R$ such that
$$
(a,b)(s,t)(a,b)^{-1}=(t,s)
$$
for every $s,t\in S$. Take $s=1$ and $t=0$.
The first question is about $\mathbb{C}$ not being central simple over $\mathbb{R}$. Suppose there are $a_k,b_k\in\mathbb{C}$ such that, for every $z\in\mathbb{C}$,
$$
\sum_{k=1}^n a_kzb_k=\bar{z}
$$
Take $z=1$ and $z=i$.
A: By commutativity of $\mathbb{C}$, the action of $M(\mathbb{C})$ on $\mathbb{C}$ is isomorphic to the action of $\mathbb{C}$ on $\mathbb{C}$ by left multiplication. In particular, with respect to an orientation of $\mathbb{C}$ as a $\mathbb{R}$-vector space, all nonzero elements of $M(\mathbb{C})$ are orientation preserving. Yet complex conjugation, viewed as an endomorphism of the real vector space $\mathbb{C}$, is orientation reversing. Hence compmlex conjugation does not belong to $M(\mathbb{C})$.
Another reason why complex conjugation is not in $M(\mathbb{C})$ is that if it were, identifying $M(\mathbb{C})$ with $\mathbb{C}$, it would have to be multiplication by $1$ (since $\overline{1}=1$), but also multiplication by $-1$ (since $\overline{i}=-i$.)

Concerning the second example, note that the units of the product ring $S\times S$ are precisely the pairs of units of $S$ :
$$U(S\times S)=U(S)\times U(S)$$
and inverses are computed coordinate wise : if $(s,t)\in U(S\times S)$, that is $s,t\in U(S)$, then
$$(s,t)^{-1}=(s^{-1},t^{-1})$$
Thus the inner automorphism associated to the unit $(s,t)$ satisfies, for all $x,y\in S$,
$$(s,t)(x,y)(s,t)^{-1}=(s,t)(x,y)(s^{-1},t^{-1})=(sxs^{-1},tyt^{-1}).$$
In particular
$$(s,t)(1,0)(s,t)^{-1}=(1,0)\quad\text{and}\quad(s,t)(0,1)(s,t)^{-1}=(0,1)$$
and hence the exchange automorphism cannot be inner.
