Yes you can extend the action. One can prove this using the universal property of blow-ups (see Hartshorne Corollary 7.15).

As $Z$ is invariant under the action of $G$, the inverse image of $Z$ with respect to the morphism $G \times X \to X$ is $G \times Z$. Therefore on applying the universal property of blow-ups to this morphism we obtain a morphism $G \times \widetilde{X} \to \widetilde{X}$. Now, by assumption this morphism satisifies the identities $$(gh)x = g(hx), \quad ex = x$$ for all $x$ in $\widetilde{X}\setminus E$, where $E$ denotes the exceptional divisor of the blow-up. However since any two morphisms which are equal on an open dense subset must be equal on the whole space, we see that these identities hold for all $x$ in $\widetilde{X}$, i.e. the morphism $G \times \widetilde{X} \to \widetilde{X}$ gives an action of $G$ on $X$.

Note that in this argument we did not use the fact that $G$ was finite, it works for any algebraic group.

Yes you can extend the action. One can prove this using the universal property of blow-ups (see Hartshorne Corollary II.7.15).

As $Z$ is invariant under the action of $G$, the inverse image of $Z$ with respect to the morphism $G \times X \to X$ is $G \times Z$. Therefore on applying the universal property of blow-ups to this morphism we obtain a morphism $G \times \widetilde{X} \to \widetilde{X}$. Now, by assumption this morphism satisifies the identities $$(gh)x = g(hx), \quad ex = x$$ for all $x$ in $\widetilde{X}\setminus E$, where $E$ denotes the exceptional divisor of the blow-up. However since any two morphisms which are equal on an open dense subset must be equal on the whole space, we see that these identities hold for all $x$ in $\widetilde{X}$, i.e. the morphism $G \times \widetilde{X} \to \widetilde{X}$ gives an action of $G$ on $X$.

Note that in this argument we did not use the fact that $G$ was finite, it works for any algebraic group.

Yes you can extend the action. One can prove this using the universal property of blow-ups (see Hartshorne Corollary 7.15).

As $Z$ is invariant under the action of $G$ on applying the universal property of blow-ups to the action morphism $G \times X \to X$ we obtain a morphism $G \times \widetilde{X} \to \widetilde{X}$. Now, by assumption this morphism satisifies the identities $$(gh)x = g(hx), \quad ex = x$$ for all $x$ in $\widetilde{X}\setminus E$, where $E$ denotes the exceptional divisor of the blow-up. However as any morphisms which are equal on an open dense subset must be equal on the whole space, we see that these identities hold for all $x$ in $\widetilde{X}$, i.e. the morphism $G \times \widetilde{X} \to \widetilde{X}$ gives an action of $G$ on $X$.

Note that in this argument we did not use the fact that $G$ was finite, it works for any algebraic group.

Yes you can extend the action. One can prove this using the universal property of blow-ups (see Hartshorne Corollary 7.15).

As $Z$ is invariant under the action of $G$, the inverse image of $Z$ with respect to the morphism $G \times X \to X$ is $G \times Z$. Therefore on applying the universal property of blow-ups to this morphism we obtain a morphism $G \times \widetilde{X} \to \widetilde{X}$. Now, by assumption this morphism satisifies the identities $$(gh)x = g(hx), \quad ex = x$$ for all $x$ in $\widetilde{X}\setminus E$, where $E$ denotes the exceptional divisor of the blow-up. However since any two morphisms which are equal on an open dense subset must be equal on the whole space, we see that these identities hold for all $x$ in $\widetilde{X}$, i.e. the morphism $G \times \widetilde{X} \to \widetilde{X}$ gives an action of $G$ on $X$.

Note that in this argument we did not use the fact that $G$ was finite, it works for any algebraic group.