Assume we have two distinct functions $f$ and $g$ such that $\phi=f^{-1}-f=g^{-1}-g$. f^{-1}-f\equiv g^{-1}-g$. Take a sequence$x_n=f(x_{n-1})$. Clearly$sign[f(x_n)-g(x_n)]=0$f(x_n)-g(x_n)=0$ or $\pm(-1)^n$. Thus, (-1)^n[f(x_n)-g(x_n)]$has the same sign for all$n$. Sinse$\int_0^1f=\int_0^1g$, there are two such sequences$x_n$and$y_n$as above such that$f(x_n)=g(x_n)$,$f(y_n)=g(y_n)$and say$(-1)^nf(x)>(-1)^ng(x)$(-1)^n[f(x)-g(x)]>0$ for any $x\in(x_n,y_n)$. Note that $x_n,y_n\to0$ x_n,y_n\to 0$and$\int_{[x_n,y_n]}|f-g|=const$. \int_{x_n}^{y_n}|f-g|=const>0$. It implies follows that $\limsup_{x\to0} |f(x)-g(x)|\to\infty$, a contradiction.
Assume we have two distinct functions $f$ and $g$ such that $\phi=f^{-1}-f=g^{-1}-g$. Take a sequence $x_n=f(x_{n-1})$. Clearly $sign[f(x_n)-g(x_n)]=0$ or $\pm(-1)^n$. Thus, there are two such sequences $x_n$ and $y_n$ such that $f(x_n)=g(x_n)$, $f(y_n)=g(y_n)$ and say $(-1)^nf(x)>(-1)^ng(x)$ for any $x\in(x_n,y_n)$. Note that $x_n,y_n\to0$ and $\int_{[x_n,y_n]}|f-g|=const$. It implies that $\limsup_{x\to0} |f(x)-g(x)|\to\infty$, a contradiction.