Denoting $_2F_1(a,b;c;z)$ as the hypergeometric function we have $$ _2F_1(a,b;c;z)=\sum_{k=0}^{\infty} \frac{(a)_{k}(b)_{k}}{(c)_{k}} \frac{z^{k}}{k !},\quad |z|<1. $$
Euler's transformation is $$ { }_{2} F_{1}(a, b ; c ; z)=(1-z)^{c-a-b}{ }_{2} F_{1}(c-a, c-b ; c ; z) $$ and two Pfaff transformations are $$ \begin{aligned} &{ }_{2} F_{1}(a, b ; c ; z)=(1-z)^{-b}{ }_{2} F_{1}\left(b, c-a ; c ; \frac{z}{z-1}\right), \\ &{ }_{2} F_{1}(a, b ; c ; z)=(1-z)^{-a}{ }_{2} F_{1}\left(a, c-b ; c ; \frac{z}{z-1}\right), \end{aligned} $$ respectively.
My question:
What is its condition for a,b,c parameters? Can c is equal a ? or a,b and c is equal zero in transformations?
Yes, these work under the conditions you ask about.
Can $a$ equal $c$?
$$ {}_2 F_1(a,b;a;z) = \sum_{k=0}^\infty \frac{(b)_k z^k}{k!} = (1-z)^{-b} \\ {}_2 F_1(0,a-b;a;z) = 1 \\ {}_2 F_1\left(b,0;a;\frac{z}{z-1}\right) = 1 \\ {}_2 F_1\left(a,a-b;a;\frac{z}{z-1}\right) = \left(1-\frac{z}{z-1}\right)^{-a+b} $$ Then your three equations become: $$ (1-z)^{-b} = (1-z)^{-b} \\ (1-z)^{-b} = (1-z)^{-b} \\ (1-z)^{-b} = (1-z)^{-a}\left(1-\frac{z}{z-1}\right)^{-a+b} $$ which are correct.
can $a, b$, or $c$ be zero?
As explained in a comment, $c$ cannot be $0$ unless one of $a,b$ is also $0$. For example $$ {}_2 F_1(1,1;0;z)\qquad\text{is no good} $$ But $$ {}_2 F_1(0,b;0;z) = (1-z)^{-b} $$ The numerator $(0)_k$ cancels the denominator $(0)_k$.