jmc

We have that $$\begin{array}{rl}f(f(x))& =f\left(\frac{ax}{x+1}\right)\\ & =\frac{a\cdot \frac{ax}{x+1}}{\frac{ax}{x+1}+1}\\ & =\frac{a^{2}x}{ax+x+1}.\end{array}$$We want $$\frac{a^{2}x}{ax+x+1}=x$$for $x\ne -1.$ This gives us $$a^{2}x=a{x}^2+x^2+x.$$Matching the coefficients, we get $a^2=1$ and $a+1=0.$ Thus, $a=\boxed{-1}.$