If you want your "reshaping" transformation function to be monotonic, then the answer is no.

Indeed, suppose that $X_j\sim\text{gamma}(a_j,1)$ for $j=1,2$. Let $F_j$ be the cdf of $X_j$.

Suppose that $X_2$ equals $f(X_1)$ in distribution for some continuous strictly increasing function $f$. Then for all real $x_2$ $$F_2(x_2)=P(X_2\le x_2)=P(f(X_1)\le x_2)=P(X_1\le f^{-1}(x_2)) =F_1(f^{-1}(x_2)),$$ so that $F_2=F_1\circ f^{-1}$ and hence necessarily $$f=F_2^{-1}\circ F_1,$$ just as you had it.

Similarly, if $X_2$ equals $f(X_1)$ in distribution for some continuous strictly decreasing function $f$, then $$f=F_2^{-1}\circ G_1,$$ where $G_1:=1-F_1$.

However, in the case when $a_1=a_2+1$, there is a curious somewhat related equidistribution phenomenon (that note can be read online for free).

If you want your "reshaping" transformation function to be monotonic, then the answer is no.

Indeed, suppose that $X_j\sim\text{gamma}(a_j,1)$ for $j=1,2$. Let $F_j$ be the cdf of $X_j$.

Suppose that $X_2$ equals $f(X_1)$ in distribution for some continuous strictly increasing function $f$. Then for all real $x_2$ $$F_2(x_2)=P(X_2\le x_2)=P(f(X_1)\le x_2)=P(X_1\le f^{-1}(x_2)) =F_1(f^{-1}(x_2)),$$ so that $F_2=F_1\circ f^{-1}$ and hence necessarily $$f=F_2^{-1}\circ F_1,$$ just as you had it.

Similarly, if $X_2$ equals $f(X_1)$ in distribution for some continuous strictly decreasing function $f$, then $$f=F_2^{-1}\circ G_1,$$ where $G_1:=1-F_1$.

However, in the case when $a_1=a_2+1$, there is a curious somewhat related equidistribution phenomenon (that note can be read online for free).

I don't think that the expression $F_2^{-1}\circ F_1$ is hard to deal with. E.g., Mathematica produces the graph $\{(x,(F_2^{-1}\circ F_1)(x))\colon0<x<7\}$ for $a_1=1.2$ and $a_2=3.4$ in about $0.052$ sec; for a comparison, it produces the graph $\{(x,\ln x)\colon0<x<7\}$ in about $0.035$ sec. Also, I think it is not hard to deal with $F_2^{-1}\circ F_1$ analytically. Here is the graph production:

If you want your "reshaping" transformation function to be monotonic, then the answer is no.

Indeed, suppose that $X_j\sim\text{gamma}(a_j,1)$ for $j=1,2$. Let $F_j$ be the cdf of $X_j$.

Suppose that $X_2$ equal $f(X_1)$ in distribution for some continuous strictly increasing function $f$. Then for all real $x_2$ $$F_2(x_2)=P(X_2\le x_2)=P(f(X_1)\le x_2)=P(X_1\le f^{-1}(x_2)) =F_1(f^{-1}(x_2)),$$ so that $F_2=F_1\circ f^{-1}$ and hence necessarily $$f=F_2^{-1}\circ F_1,$$ just as you had it.

Similarly, if $X_2$ equal $f(X_1)$ in distribution for some continuous strictly decreasing function $f$, then $$f=F_2^{-1}\circ G_1,$$ where $G_1:=1-F_1$.

If you want your "reshaping" transformation function to be monotonic, then the answer is no.

Indeed, suppose that $X_j\sim\text{gamma}(a_j,1)$ for $j=1,2$. Let $F_j$ be the cdf of $X_j$.

Suppose that $X_2$ equals $f(X_1)$ in distribution for some continuous strictly increasing function $f$. Then for all real $x_2$ $$F_2(x_2)=P(X_2\le x_2)=P(f(X_1)\le x_2)=P(X_1\le f^{-1}(x_2)) =F_1(f^{-1}(x_2)),$$ so that $F_2=F_1\circ f^{-1}$ and hence necessarily $$f=F_2^{-1}\circ F_1,$$ just as you had it.

Similarly, if $X_2$ equals $f(X_1)$ in distribution for some continuous strictly decreasing function $f$, then $$f=F_2^{-1}\circ G_1,$$ where $G_1:=1-F_1$.

However, in the case when $a_1=a_2+1$, there is a curious somewhat related equidistribution phenomenon (that note can be read online for free).