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. 

Consider the standard Euclidean vector space $\mathbb{R}^1$ and let $F : \mathbb{R}^1 \to \mathbb{R}^1$ be the non-linear map defined as $F(x) = x + 1$. Even though $F(0) = 1 \neq 0$, this map is an isometry, because we have
$$
\lvert F(x) - F(y) \rvert
= \lvert (x+1) - (y+1) \rvert
= \lvert x - y \rvert \;.
$$
