Let $A$ be an algebra and let $A^{\ast}$ be the subset of units (that is, invertible elements) of $A$. Then $A^{\ast}$ is a group under the multiplication of $A$.
Let $f^{\ast}:A^{\ast}\to A^{\ast}$ be a group isomorphism. Is it true that there always exists an isomorphism (between algebras; so it preserves addition and multiplication structure) $f:A\to A$ such that $f|_{A^{\ast}}=f^{\ast}$? If yes, is such $f$ unique?
On
Consider the $\mathbb{R}$ algebra $\mathbb{R}$ the group of units $U=\mathbb{R}^*$. Let us take the automorphism of $U$ which send $x $ to $\frac{1}{x}$. Now this does not extend to an algebra automorphism of $\mathbb{R}$ since any such morphism will send 1 to 1 and hence will fix the rationals. Where in this map the non-zero rationals are sent to it's inverse.
First this looks like a statement on rings rather than a statement over algebras. Anyway, take $A$ to be the field with $5$ elements. Then $A^*$ is a cyclic group with $4$ elements, it admits one non-trivial automorphism $f^*$. However if $f$ is an automorphism of ring then it must $1$ to $1$ but this implies that $f$ is the identity. Hence no ring automorphism can extend $f^*$.