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.