I'm fairly sure the following is true, although I wouldn't mind being proven wrong. If true, I would like to see an elegant proof, as my attempts are kind of messy.
Let $K\subset\mathbb R^2$ be a compact convex set. Let $x_0\in K$ be an extreme point. The question is whether there exists an affine transformation $\phi:\mathbb R^2\to\mathbb R^2$ (i.e., $\phi(x)=r+Bx$ for $r\in\mathbb R^2$ and $B\in M_2(\mathbb R)$) such that $\forall y\in\phi(K)\setminus\{\phi(x_0)\}$, $$ \|y\|_2<\|\phi(x_0)\|_2.$$
In words, the question is whether we can rotate and translate $K$ in such a way that $x_0$ ends up being strictly farthest from the origin. As Tryss mentioned below, another way to phrase the problem is to ask whether a circle $C$ exists with $x_0\in C$ and $K\setminus\{x_0\}\subset C$ strictly.
I think I have a counterexample. Define $$K = \Big\{(x,y) \in \big[-\frac12,\frac12\big] \times \big[0, \frac12\big] : y \ge \exp\big(-\frac1{x^2}\big) \Big\}.$$
Now, consider the extreme point $x_0 = (0,0)$. The key observation is that all derivatives of $x \mapsto f(x) := \exp\big(-\frac1{x^2}\big)$ vanish at $0$, hence, all circles (ellipses) containing $x_0$ will also intersect with the graph of $f$.
Or speaking in terms of curvature: the curvature of $K$ at $x_0$ is $0$, although $x_0$ is an extreme point. If the curvature would be positive, we could use the osculating circle, as pointed out by Tryss.