jmc

Substituting $f^{-1}(x)$ into our expression for $f$, we get $$f(f^{-1}(x))=5f^{-1}(x)-12.$$Since $f(f^{-1}(x))=x$ for all $x$ in the domain of $f^{-1}$, we have $$x=5f^{-1}(x)-12.$$Solving for $f^{-1}(x)$ gives $$f^{-1}(x)=\frac{x+12}{5}.$$The equation $f^{-1}(x)=f(x+1)$ now reads $$\frac{x+12}{5}=5(x+1)-12=5x-7.$$Cross-multiplication gives $$x+12=25x-35.$$Isolating $x$ gives us $$24x=47.$$Solving for $x$, we find $x=\boxed{\frac{47}{24}}$.