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Gregor Kemper answered a related question with a technique that can be used to answer this one affirmatively in the case that $g$ has finite order. If $g$ does not have finite order, the answer is negative.

Gregor Kemper answered a related question with a technique that can be used to answer this one affirmatively in the case that $g$ has finite order. If $g$ does not have finite order and we drop the integrality assumption, the answer is negative.

If $g$'s restriction to $B$ has infinite order, then its action on $B$ may not be linearizable.

The previous argument fails in this situation because the conclusion of Maschke's theorem fails: while $I^2$ is still a $g$-invariant subspace of $I$, it does not have a $g$-invariant complement.

If $g$'s restriction to $B$ has infinite order, and we drop the hypothesis that $A$ is integral over $B$, then $g$'s action on $B$ may not be linearizable.

The previous argument fails in this situation because the conclusion of Maschke's theorem fails: while $I^2$ is still a $g$-invariant subspace of $I$, it does not have a $g$-invariant complement.

But note that in this situation, $A$ is not integral over $B$, as $y$ is not integral over $B$.

I do not know if there is an example of infinite order $g$ and $A/B$ integral in which $g|_B$ is not linearizable.

Gregor Kemper answered a related question with a technique that can be used to answer this one in the case that $g$ has finite order.

I do not know if the result holds in characteristic 0 case when $g|_B$ has infinite order. Evidently, this argument does not go through, since the conclusion of Maschke's theorem doesn't hold in general. But I do not have an example of a $B$ and $g$ as in the OP for which $I^2$ does not have a $g$-invariant complement in $I$.

Gregor Kemper answered a related question with a technique that can be used to answer this one affirmatively in the case that $g$ has finite order. If $g$ does not have finite order, the answer is negative.

If $g$'s restriction to $B$ has infinite order, then its action on $B$ may not be linearizable.

Let $A=\mathbb{C}[x,y]$ and let $B=\mathbb{C}[x,xy]$. Let $g$ act on $A$ by $x\mapsto x$, $y\mapsto x+y$. Then $gB\subset B$ since $xy\mapsto x^2+xy\in B$, and $B\subset gB$ since $xy\in \mathbb{C}[x,x^2+xy]=gB$. Thus $B$ is fixed setwise by $G$. But $B$ is algebra-generated in distinct degrees and $g|_B$ does not act diagonally on the generators, so it is not linearizable.

The previous argument fails in this situation because the conclusion of Maschke's theorem fails: while $I^2$ is still a $g$-invariant subspace of $I$, it does not have a $g$-invariant complement.