Given a function $F(z)$, the function $F(z-m)$ is called the Taylor shift of $F(z)$.
If $F(z)$ is a (general) hypergeometric function, is $F(z-m)$ some simple transformation of a hypergeometric function?
Typical transforms of $F(z)$ are $A(z) F(B(z))$ where $A(z), B(z)$ are polynomials.

More precisely, suppose $F(z)={}_pF_q((a);(b);z)$ where $(a),(b)$ are the parameters, is $F(z-m)$ is some simple transformation of another hypergeometric function. But the Taylor shift is already in this form: choose $A(z)=1, B(z)=x-m$. Call this the trivial transformation. My question is whether non-trivial transformations always exist.

For example, $F(z)={}_1F_0(a;;z)$, I can show the non-trivial transformation, $$ F(z-m) = (1+m)^{-a} F\big(\frac{z}{1+m}\big) $$

But does the Taylor shift of $F(z)={}_0F_1(;b;z)$ have a nontrivial transformation?

Taylor shift is very important for efficient algorithms. Also, any pointers or references would be appreciated. This is my first question in mathoverflow. Thanks to the editor who fixed my latex (removing indents).