Edited after Mark's comment

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), then it follows that $g'=g$ and so $f'$ is the required lift. Functorial resolutions exist by 3.36 of Kollár.

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.

Take a resolution $g:Y\to X$ and consider the induced maps from $f, f^{-1},f$: $$ X\times _X Y \to Y \to X\times _X Y \to Y, $$ and conclude that these are isomorphisms as well.

Edited after Mark's comment

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), then it follows that $g'=g$ and so $f'$ is the required lift. Functorial resolutions exist by 3.36 of Kollár.