Does "F(x) is isometric" imply "F(x) is linear"? I'm asked to prove that if $F(x)$ is isometric and satisfies $F(0) = 0$ then $F(x)$ must be linear. 
I've got a pretty rigorous proof that "$F(x)$ is isometric" $\implies$ "$F(x)$ is linear" and thus the condition $F(0) = 0$ need not be satisfied. (My "proof" however assumes that angles don't change under a isometric transformation, but this seems reasonable). 
My Question: Why is the condition $F(0) = 0$ needed? I realize that if that condition is satisfied, then $F(x)$ is simply a rotation in 2D. But I can't find any isometric transformation where $F(0) \neq 0 $ which isn't linear. 


(My "proof" however assumes that angles don't change under a isometric transformation, but this seems reasonable).

This is correct. Being an isometry is a very strong condition. If $F$ is an isometry, then $F$ also preserves angles. For an example in 2D, we can see that $F$ sends $\Delta ABC$ to a congruent triangle $\Delta A'B'C'$:

In the picture, the mapping $F$ is a reflection across the line $DE$. 
