= Solution
A <Möbius transformation of a Papperitz symbol> changes the independent variable by a <Möbius transformation>. In particular,
$$
u=\frac{(b-c)(z-a)}{(b-a)(z-c)}
$$
maps $a,b,c$ to $0,1,\infty$, respectively. Because a Möbius map is locally biholomorphic away from its pole, it carries each exponent pair with its singular point and gives
$$
P\left\{
\begin{matrix}
0&1&\infty\\
\alpha&\beta&\gamma\\
\alpha'&\beta'&\gamma'
\end{matrix};u
\right\}.
$$
Now define a <dependent-variable rescaling of a Papperitz symbol> by writing the old dependent variable as
$$
y(u)=u^\alpha(1-u)^\beta v(u),
$$
so that $v=u^{-\alpha}(1-u)^{-\beta}y$. This <dependent-variable rescaling of a Papperitz symbol> subtracts $\alpha$ from both exponents at zero, subtracts $\beta$ from both exponents at one, and adds $\alpha+\beta$ to both exponents at infinity. Hence $v$ has symbol
$$
P\left\{
\begin{matrix}
0&1&\infty\\
0&0&\gamma+\alpha+\beta\\
\alpha'-\alpha&\beta'-\beta&\gamma'+\alpha+\beta
\end{matrix};u
\right\}.
$$
Define
$$
A=\gamma+\alpha+\beta,
\qquad
B=\gamma'+\alpha+\beta,
\qquad
C=1+\alpha-\alpha'.
$$
The Fuchs relation gives $\alpha'-\alpha=1-C$ and $\beta'-\beta=C-A-B$. After renaming $u$ as $z$, the symbol is therefore
$$
\boxed{
P\left\{
\begin{matrix}
0&1&\infty\\
0&0&A\\
1-C&C-A-B&B
\end{matrix};z
\right\}},
$$
which is the <Gauss hypergeometric equation>. Renaming $A,B,C$ as $a,b,c$ gives exactly the displayed symbol in the question.
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