Picard-surjectivity and Morita-equivalence Let us say that an algebra $A$ over a field $k$ is Picard-surjective if the canonical map
$$ \mathrm{Aut}(A) \rightarrow \mathrm{Pic}(A)$$
is surjective. Here $\mathrm{Pic}(A)$ denotes the group of isomorphism classes of invertible $A$-$A$-bimodules and the map sends an automorphism $\alpha$ to the $A$-$A$-bimodule $A_\alpha$, where the left action is the usual one and the right action is via $\alpha$.
Q: For any given finite-dimensional $k$-algebra $A$, does there exist a Morita-equivalent one that is Picard-surjective?
If not, I am interested in conditions under which this is true. I am mainly interested in the case $k=\mathbb{R}$ or $\mathbb{C}$, and for all the examples that I came up with so far, this seems to be correct, as far as I can tell.
 A: Yes, the basic algebra of $A$ will be Picard-surjective.
The basic algebra is the endomorphism algebra $\operatorname{End}_A(\bigoplus_{i=1}^{n}P_i)$ of the direct sum of indecomposable projective (right) modules, one from each isomorphism class. It is Morita equivalent to $A$.
Suppose $A$ is basic. Then as a left or right module, $A$ is the direct sum of indecomposable projective modules, one from each isomorphism class. Suppose $M$ is an invertible bimodule. Since $S\otimes_AM$ is nonzero for every simple module $S$, a direct sum decomposition of $M$ as a left module must contain at least one copy of each indecomposable projective. Let $X=\bigoplus_{i=1}^nS_i$ be the direct sum of simple (right) $A$-modules, one from each isomorphism class. Since $X\otimes_AM\cong X\cong X\otimes_AA$, as a left module $M$ must contain exactly one copy of each indecomposable projective in a direct sum decomposition. So $M\cong A$ as left modules. The right $A$-module structure of $M$ is then given by an injective algebra homomorphism $A^{op}\to\operatorname{End}_A(_AA)\cong A^{op}$, which is an isomorphism by finite dimensionality: i.e., the right action is induced by an automorphism of $A$.
