Let $k=\mathbb{R}$ or $\mathbb{C}$ and let $A$ be a finite-dimensional $k$-algebra. If $A$ is simple, then the Skolem-Noether theorem says that any two algebra homomorphisms $f, g: A \to M\_n(k)$ are conjugate in $M\_n(k)$, i.e., there exists a matrix $X \in M\_n(k)$ such that for all $a \in A$, $f(a) = X g(a) X^{-1}$.


It we do not assume that $A$ is simple, then this result is generally false (see e.g. [the example given by Denis Serre](https://mathoverflow.net/a/84600) at [Conjugation between commutative subalgebras of a matrix algebra?](https://mathoverflow.net/questions/84591/conjugation-between-commutative-subalgebras-of-a-matrix-algebra)).


**My question is:** Is this result still true if $f$ and $g$ are "close"? That is, if $f$ and $g$ are close, then they are conjugate?


Here by "close" I mean that for some small $\varepsilon>0$, we have $\lVert f(a) - g(a)\rVert \leq \varepsilon \lVert a\rVert$, for some submultiplicative norm on $A$.



Are nearby subalgebras of matrix algebras conjugate?
I'll describe a 1-parameter family of nonisomorphic 4-dimensional subalgebras of $M\_4(K)$. Consider, for $t\in K^\*$, the matrices
$$X=\begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix},\;Y\_t=\begin{pmatrix} 0 & 1 & 1 & 0\\ 0 & 0 & 0 & t\\ 0 & 0 & 0 & -t \\ 0 & 0 & 0 & 0 \end{pmatrix},\;Z=\begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$
Then $XY\_t=tZ$, $Y\_tX=Z$, and all other pairwise products (including squares) between these matrices are zero. Hence they form the basis of a 3-dimensional (non-unital) subalgebra $N\_t$. These algebras are pairwise non-isomorphic, except $N\_t\simeq N\_{t^{-1}}$ (indeed, in $N\_t$, the system of equations $$\{xy=\lambda yx;\quad x^2=y^2=0\}$$ has a solution $(x,y)$ with $xy\neq 0$ iff $\lambda\in\{1,t,1/t\}$).


Let now $A\_t$ be the 4-dimensional unital subalgebra with basis $(I,X,Y\_t,Z)$. These are also pairwise nonisomorphic (except $t\leftrightarrow t^{-1}$), since $N\_t$ consists of the set of nilpotent elements in $A\_t$.


Since they are not isomorphic, they are not conjugate.


Now let $A$ be 5-dimensional, with basis $(I,X,Y,Z,Z')$ with $I$ identity, $XY=Z'$, $YX=Z$, and all other products (not involving $I$) being zero. Let $f\_t:A\to A\_t\subset M\_4(K)$ map $I\mapsto I$, $X\mapsto X$, $Y\mapsto Y\_t$, $Z\mapsto Z$, $Z'\mapsto tZ$. Then $f\_t$ is a surjective homomorphism. However they have non-isomorphic images (up to $t\leftrightarrow t^{-1}$). While for $K$ being real or complex numbers, for $t'$ close enough to $t$, the homomorphisms $f\_t$ and $f\_{t'}$ are close.


