Arrow's Theorem for Cardinal Utility

Say that utility functions $u_n$ and $v_n$ are ordinally equivalent if $v_n=\phi\circ u_n$ where $\phi$ is increasing, and cardinally equivalent if $\phi$ is a positive affine transformation. It suffices to show that any two ordinally equivalent utility profiles are mapped to the same social preference relation. The axioms then imply that the conditions for Arrow's theorem are met, which proves the theorem.

Choose $x,y$ in $X$, and utility profile $u$. Let $v$ be another profile in which each each $v_n$ is ordinally equivalent to $u_n$. To show that the social preference between $x$ and $y$ must be the same, find a profile $u'$ cardinally equivalent profile to $u$ such that for each $n$, $u'_n(x)=v_n(x)$ and $u'_n(y)=v_n(y)$. By cardinality, the social preference for $x$ versus $y$ for $F(u')$ and $F(u)$ must agree, and for $F(u')$ and $F(v)$ because of IIA. Thus $F(u)$ and $F(v)$ agree on the $x,y$ comparison.

If $u_n(x)=u_n(y)$ then $u_n'(z)=(v_n(x)-u_n(x))+u_n(z)$ equals $u_n'(x)$ for $z=x,y$. If $u_n(x)>u_n(y)$, take $u_n'(z)=a+bu_n(z)$ where $b=(v_n(x)-v_n(y))/(u_n(x)-u_n(y))>0$ and $a=(u_n(x)v_n(y)-u_n(y)v_n(x))/(u_n(x)-u_n(y)$. Again, $u'_n(z)=v_n(z)$ for $z=x,y$. ∎