Suppose I have a commutative ring $R$ and a finite-dimensional $R$-algebra $A$, and a so-called basic idempotent $e$ (an idempotent $e$ such that $eAe$ is a basic algebra of $A$). It is known that $A$ is Morita equivalent to $eAe$, so in particular $\operatorname{Pic}(A)=\operatorname{Pic}(eAe)$ (the Picard group).
If we consider the group of outer automorphisms, it is known that there is an injective map $\operatorname{Out}(A) \to \operatorname{Pic}(A)$, and actually for any basic algebra $B$ it holds that $\operatorname{Out}(B)=\operatorname{Pic}(B)$.
So, my question: given how $\operatorname{Out}(A)$ is "smaller" than $\operatorname{Out}(eAe)$, I expect that given an outer automorphism of $A$ I can somehow get an outer automorphism of $eAe$. However, I cannot find a "good" way to do this, since given a $\phi \in \operatorname{Out}(A)$ then maybe $\phi(e) \neq e$, so I cannot restrict $\phi$ to $eAe$.
On the other hand, it seems that if I fix a basic idempotent $e$ and then I consider $$\operatorname{Out}(A) \to \operatorname{Pic}(A) \simeq \operatorname{Pic}(eAe) = \operatorname{Out}(eAe)$$ I should get a well-defined $\tilde{\phi} \in \operatorname{Out}(eAe)$. But I have trouble working with it, because $\phi(e) \neq e$. What am I missing?
It is known that over a semiperfect ring $A$, any two basic idempotents are conjugate. For example, this is Exercise 25.4 in T.Y.Lam's Exercises in Classical Ring Theory. Thus we can find a unit $u\in A$ such that $\gamma_u\phi(e)=e$, where $\gamma_u\colon x\mapsto uxu^{-1}$ is conjugation by $u$. Since $\phi$ and $\gamma_u\phi$ determine the same outer automorphism, this gives the required map.