**The First question**
Let $A$ be a Banach or a $C^\*$ algebra. Assume that $e,f$ are two idempotents or prjections in $A$ which satisfy $ef=fe=0$. Assume that there are two automorphisms $\phi, \psi: A \to A$ with $\phi(e)\psi(f)=\psi(f)\phi(e)=0$.



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> 1) Is there an automorphism $\rho:A\to A$ with $\rho(e)=\phi(e),\;\rho(f)=\psi(f)$?
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> 2) **A topological version:** Let $X$ be a compact Hausdorff space and $B, C$ be two disjoint **Clopen** subsets of $X$. Assume that $F, G$ are two homeomorphisms of $X$ with $F(B)\cap G(C)= \varnothing$.Is there a homeomorphism $H$ of $X$ with $H(B)=F(B),\;H(C)=G(C)$?
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**The second question:**
Let $A$ be a (Banach or a $C^\*$ ) algebra or a ring. Assume that $e,f$ are two idempotents or projections in $A$. Assume that $\phi,\psi A:\to A$ are two automorphisms. Is there an automorphism $\alpha$ of $M\_2(A)$ with $ \alpha (e\oplus f)=\phi(e)\oplus \psi(f)$?


**Note:** A positive answer to these questions enable us to stablish a group structure on a possible [new kind of K-theory of rings or algebras](https://mathoverflow.net/questions/348508/other-kinds-of-equivalence-relations-on-the-set-of-idempotents-of-a-banach-or-c)



Automorphism of algebras with certain initial conditions on given idempotents
**Edit:** The answer to both of your questions is "no". 


The counterexamples below all carry a similar flavor.




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**Counterexample to the topological version of the first question.** (Taking the $C^\*$-algebra of continuous $\mathbb{C}$-valued functions on the counterexample gives a counterexample to the $C^\*$-algebra version.)


Let $X$ be the following subset of $\mathbb{R}$ with the subspace topology:
$$
X=\{-2,0,2\}\cup\{2-\frac{1}{n}\,|\,n\in\mathbb{N}\}\cup\{-2+\frac{1}{n}\,|\,n\in\mathbb{N}\}.
$$
There is a homeomorphism $F:X\to X$ fixing $-2$ and $2$ and taking each $x\in X-\{\pm 2\}$ to $\min (X\cap (x,2])$ (i.e. the closest point on the right).


Let $B=\{2-\frac{1}{n}\,|\,n=1,2,\dots\}\cup\{2\}$ and $C=\{-2+\frac{1}{n}\,|\,n=1,2,\dots\}\cup\{-2\}$ and write $G=F^{-1}$. 


Then
$F(B)=\{2-\frac{1}{n}\,|\,n=2,3,\dots\}\cup\{2\}$ and $G(C)=\{-2+\frac{1}{n}\,|\,n=2,3,\dots\}\cup\{-2\}$
are disjoint, but there is no homeomorphism $H:X\to X$ taking $B$ to $F(B)$ and $C$ to $G(C)$. Indeed, if $H$ existed, then it would restrict to a bijection between $X-B-C=\{0\}$ and $X-F(B)-G(C)=\{-1,0,1\}$, which is absurd.




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**Counterexample to the the second question and the $C^\*$-algebra version of the first question.**


Take $A$ to be the abstract ring $\mathbb{C}^{\mathbb{Z}}$. Alternatively, take the subring of $\mathbb{C}^{\mathbb{Z}}$ consisting of sequences $(a\_n)\_{n\in \mathbb{Z}}$ so that both $\lim\_{n\to \infty }a\_n$ and $\lim\_{n\to -\infty}a\_n$ exist --- this is a $C^\*$-algebra which is isomorphic to $C(X)$ with $X$ as above.


Let $e\_s$ denote the sequence $(a\_n)\_{n\in \mathbb{Z}}$ for which $a\_n=0$ if $n<s$ and $a\_n=1$ if $n\geq s$, and let $f\_s=1\_A-e\_s$.


It is easy to see that there are automorphisms of $A$ carrying $e\_0$ to $e\_1$ and $f\_0$ to $f\_{-1}$. However, there is not automorphism of $A$ carrying $e\_0$ to $e\_1$ and $f\_0$ to $f\_{-1}$ because $e\_0+f\_0=1\_A$ while $e\_1+f\_{-1}\neq 1\_A$.
Furthermore, there is no automorphism of $\mathrm{M}\_2(A)$ taking $e\_0\oplus f\_0$ to $e\_1\oplus f\_{-1}$, because $\mathrm{Ann}\_{\mathrm{Cent}(\mathrm{M}\_2(A))}(e\_0\oplus f\_0)=0$ while $\mathrm{Ann}\_{\mathrm{Cent}(\mathrm{M}\_2(A))}(e\_1\oplus f\_{-1})$ is nonzero.


