Given a function $F(z)$, the function $F(z-m)$ is called
the Taylor shift of $F(z)$.<br> If $F(z)$ is a (general) hypergeometric
function, is $F(z-m)$ some simple transformation of a hypergeometric
function?<br>  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).