Automorphisms of singular varieties If $f : X \to X$ is an automorphism of a singular variety over $\mathbb C$, does there exist a resolution of singularities $g : Y \to X$ such that the induced birational map on $Y$ is also an automorphism?
 A: Edited after Mark's comment. Also after ACL's comment. Cheers.
Take a resolution $g:Y\to X$ and 
let $Y'=X\times_{f,g} Y$ with projection maps $f':Y'\to Y$ and $g':Y'\to X$. Next let $Y''=X\times_{f^{-1},g'} Y'$ with projection maps $f'':Y''\to Y'$ and $g'':Y''\to X$ and $Y'''=X\times_{f,g''} Y''$ with projection maps $f''':Y'''\to Y''$ and $g''':Y'''\to X$.
Observe that $Y''=X\times_{f^{-1},g'} Y'= X\times_{f^{-1},g'}(X\times_{f,g} Y)\simeq X\times_{\mathrm{id}_X ,g}Y\simeq Y$ where the last isomorphism is given by $f'\circ f''$ and also, clearly, $g''=g$.
Similarly $Y'''=X\times_{f,g''} Y''\simeq X\times_{f,g'}(X\times_{f^{-1},g} Y')\simeq X\times_{\mathrm{id}_X ,g'}Y'\simeq Y'$  where the last isomorphism is given by $f''\circ f'''$ and also, clearly, $g'''=g'$ and then $f'''=f'$.
So we get that for an arbitrary resolution $g:Y\to X$ there is another resolution $g':Y'\to X$ such that $f':Y'\to Y$ is an isomorphism with inverse $f'':Y\to Y''$.
Now if we choose $g:Y\to X$ to be a functorial resolution of $X$ (as in 3.4 of Kollár's book), then it follows that $g'=g$ and so $f'$ is the required lift. Functorial resolutions exist by  3.36 of Kollár's book.
