Imagine that you have a vector field $A = \frac{A_0}{r} e_{\theta}$ in cylindrical coordinates, where $A_0 \in \mathbb{R}$. Now you translate your coordinate system in $e_x$ direction by $x \mapsto x + d$ for some constant $d \in \mathbb{R}$. What happens to the field then?
I would say that $\frac{A_0}{r} \mapsto \frac{A_0}{\sqrt{(x-d)^2+y^2}}$, but what happens to the unit vector $e_{\theta}$?
So the question is: How can we express this vector field in the shifted coordinate system? If anything is unclear, please let me know.
The coordinate change you wish to study is most natural in Cartesian terms. Therefore, change the given $A$ to standard Cartesians as a starting point: $$ e_{\theta} = \langle -\sin \theta, \cos \theta \rangle = \langle \frac{-y}{r}, \frac{x}{r} \rangle $$ Therefore, $$ A = \frac{A_o}{r}e_{\theta} = \frac{A_o}{x^2+y^2} \langle -y, x \rangle $$ Let $x' = x+d$ and $y' = y$ then $x = x'-d$ and $y = y'$. Thus, $$ A = \frac{A_o}{r}e_{\theta} = \frac{A_o}{(x'-d)^2+(y')^2} \langle -y', x'-d \rangle $$ In the prime coordinates we also introduce $r', \theta'$ where these are defined implicitly by $$ x' = r' \cos \theta', \qquad y' = r'\sin \theta'$$ hence $r' = \sqrt{(x')^2+(y')^2}$ and $\tan \theta' = y'/x'$. Returning to $A$ we find, $$ A=\frac{A_o}{(r' \cos \theta'-d)^2+(r'\sin \theta')^2} \langle -r'\sin \theta', r' \cos \theta'-d \rangle $$ I suppose you probably want the end result in terms of the prime-polar frame $e_{r'}, e_{\theta'}$. Note $\nabla x' = \nabla x$ and $\nabla y' = \nabla y$ hence $e_x=e_{x'}$ and $e_y=e_{y'}$. Therefore, $$ A=\frac{A_o}{(r' \cos \theta'-d)^2+(r'\sin \theta')^2} \left( -r'\sin \theta'e_{x'}+ (r' \cos \theta'-d)e_{y'} \right) $$ Almost there, now just convert $e_x=e_{x'}$ and $e_y=e_{y'}$ to the prime polar frame. The cartesian and polar prime frames are related in the usual manner: $$ e_{r'} = \cos \theta' e_{x'}+ \sin \theta' e_{y'} \qquad e_{\theta'} = -\sin \theta' e_{x'}+ \cos \theta' e_{y'} $$ which inverted yield $$ e_{x'} = \cos \theta' e_{r'}-\sin \theta' e_{\theta'} \qquad e_{y'} = \sin \theta' e_{r'}+\cos \theta' e_{\theta'}$$ Once more, return to $A$, $$ A=\tfrac{A_o}{(r' \cos \theta'-d)^2+(r'\sin \theta')^2} \left( -r'\sin \theta'[\cos \theta' e_{r'}-\sin \theta' e_{\theta'}]+ (r' \cos \theta'-d)[\sin \theta' e_{r'}+\cos \theta' e_{\theta'}] \right) $$ which simplifies nicely to: $$ A=\tfrac{A_o}{(r' \cos \theta'-d)^2+(r'\sin \theta')^2} \left[ (r'-d\cos \theta ') e_{\theta'}-d\sin \theta' e_{r'} \right] $$ which reminds me of a dipole field formula from my junior level electromagnetics course.