Let's start with u=z/10, k=1.9/(1+n). We want the function to be just x when u=1, and something like exp(k(x-1)) when u=0, but the interpolation has to be nonlinear, because of the middle graph you show when n=+inf. If it's linear, you can just do ux+(1-u)exp(k(x-1)), or `zx/10+(1-z/10)exp(1.9(x-1)/(1+n))`.

If you imagine rotating the functions 180 degrees and pushing them up and to the right by 1 unit, they look like x and 1-exp(-kx), it's probably clearer to think that way.

Judging from your graphs, one idea is to ensure linear interpolation on the left end, something else on the right end. Call g=exp(k(x-1)) for simplicity, then perhaps ux+(1-u)g when x=-1, (u+1-(1-u)^(m))x+(1-u+(1-u)^(m))g when x=1, for some m like 2. That can be done by replacing the exponent m with m+1+x(m-1))/2. The final function would look like `(z/10+1-(1-z/10)^((3+x)/2))x+(1-z/10+(1-z/10)^((3+x)/2))exp(1.9(x-1)/(1+n))`. A problem with that is that sometimes but especially when n<1, the low-z functions will be lower than the high-z ones. Replacing 1+n with 2+n mostly fixes that, but feels dirty, and it's probably always problem for a range of x values sufficiently close to 1. EDIT: divide those by 2.

Another idea is to use hyperbolae (or conic sections in general) instead of an exponential function. Solving an implicit equation like (x+a)(y+b)c+((x+a)^(2)-(y+b)^(2))sqrt(1-c^(2))=d, finding a way to get the tangent at x=1,y=1 with slope 1, except in the z=0,n=+inf case, and adjusting the other parameters based on z and n to get the right curve. Probably overcomplicates things.