If the Hom-space of finite length modules is generated by single elements, must the elements be conjugate?

Question

Let $A$ be an Artin $k$-algebra for a commutative artinian ring $k$ (e.g. $A$ is a finite dimensional algebra over a field $k$). Let $X,Y$ be finite length left $A$-modules. If $\text{Hom}_A(X,Y)$ is generated by $f$ (respectively $g$) as an $\text{End}_A(Y)$-$\text{End}_A(X)^\text{op}$-bimodule, does it follow that $f = \beta g \alpha$ for isomorphisms $\alpha \colon X \rightarrow X$ and $\beta \colon Y \rightarrow Y$? This seems to be the case in every example that I have considered.

EDIT: The statement as above is wrong, see the answer of Dave Benson. Instead of requiring that $\text{Hom}_A(X,Y)$ is generated by $f$ (respectively $g$) as an $\text{End}_A(Y)$-$\text{End}_A(X)^\text{op}$-bimodule, I actually want to require that every element in $\text{Hom}_A(X,Y)$ is of the form $f\alpha'+\beta'f$ for some $\alpha' \colon X \rightarrow X$ and $\beta' \colon Y \rightarrow Y$ (same for $g$). Does the implication hold in this case?

Answer 1

This isn't even true for vector spaces over a field. Let $A=k$ be a field. Let $X$ and $Y$ be two dimensional vector spaces, and let $f$ be the homomorphism $\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right)$ and let $g$ be the identity matrix, as homomorphisms from $X$ to $Y$. Then $f$ generates $\operatorname{\rm Hom}_k(X,Y)$ as a bimodule over the endomorphism rings, since this bimodule is simple. Similarly for $g$. If there were such $\alpha$ and $\beta$ then $f$ would be an isomorphism.