$\def\RR{\mathbb{R}}$
The answer is no! I will give an example of a centrally symmetric polytope in $\RR^4$ with $12$ vertices where there is a symmetry of the edge graph interchanging two non-antipodal vertices, and fixing the other ten vertices.
Define the function $f : \RR \to \RR^4$ by
$$f(\theta) = (\cos \theta, \sin \theta, \cos (3 \theta), \sin (3 \theta)).$$
Observe that $f(\theta+\pi) = -f(\theta)$, so the convex hull of $f(\theta_1)$, $f(\theta_2)$, ..., $f(\theta_n)$, $f(\theta_1+\pi)$, $f(\theta_2+\pi)$, ..., $f(\theta_n+\pi)$ is always a centrally symmetric polytope.
I'll write $P(\theta_1, \theta_2, \ldots, \theta_n)$ for the convex hull of $f(\theta_1)$, $f(\theta_2)$, ..., $f(\theta_n)$, $f(\theta_1+\pi)$, $f(\theta_2+\pi)$, ..., $f(\theta_n+\pi)$.
We need two lemmas:
Lemma 1: Let $|\theta_1 - \theta_2| < 2 \pi/3$ and let $\theta_3$, $\theta_4$, ..., $\theta_n$ be any other angles. Then $(f(\theta_1), f(\theta_2))$ is an edge of $P(\theta_1, \theta_2, \theta_3, \theta_4, \ldots, \theta_n)$.
Lemma 2: Let $0 < \alpha < \beta < \pi/2$. Then there is $\delta>0$ (dependent on $\alpha$ and $\beta$) such that, for $|\gamma-\pi/2| < \delta$, the line segement $(f(\gamma), f(- \gamma))$ is NOT an edge of $P(-\gamma, -\beta, - \alpha, \alpha, \beta, \gamma)$.
Once we have these lemmas, our construction will be to choose $0 < \alpha < \beta < \pi/6$ and then $\gamma$ extremely close to $\pi/2$.
Our polytope will be $P(-\gamma, - \beta, - \alpha, \alpha, \beta, \gamma)$. If we have chosen $\gamma$ close enough to $\pi/2$, then the above lemmas guarantee that both $f(\gamma)$ and $f(\pi - \gamma)$ will NOT neighbor $f(- \gamma)$ and $f(-\pi+\gamma)$, but will neighbor the other eight vertices (and each other). So switching $f(\gamma)$ and $f(\pi-\gamma)$ will be an symmetry of the edge graph which does not preserve the antipodal pairing.
We now prove the lemmas.
Proof of Lemma 1: Rotating the circle, we may assume that $\theta_1 = - \theta_2$ and $0 < \theta_1 < \pi/3$. Put $a = \cos \theta_1 > 1/2$ and consider the function $g(\theta) = 3 a^2 \cos \theta - \cos^3 \theta$. Basic calculus shows that this is maximized at $\theta = \pm \theta_1$. (We need that $a>1/2$ in order to make sure that the value at $\theta_1$, namely $2 a^3$, beats the other local maximum at $\pi$, namely $1-3a^2$.) Expanding $\cos^3 \theta = (3/4) \cos \theta + (1/4) \cos (3 \theta)$, we have $g(\theta) = (3 a^2-3/4) \cos \theta - (1/4) \cos (3 \theta)$. Writing $(x_1, x_2, x_3, x_4)$ for the coordinates on $\RR^4$, the linear functional $(3a^2 - 3/4) x_1 - (1/4) x_3$ is larger at $f(\pm \theta_1)$ than at any other $f(\theta)$, so $((f(\theta_1), f(-\theta_1))$ is an edge of
$P(\theta_1, -\theta_1, \theta_3, \theta_4, \ldots, \theta_n)$ as desired. $\square$
Proof of Lemma 2: It is enough to show that some point on the line segment from $f(\gamma)$ to $f(- \gamma)$ is in the convex hull of $f(\pm \alpha)$, $f(\pm \beta)$, $f(\pi \pm \alpha)$ and $f(\pi \pm \beta)$.
Putting $h(\theta) = ((f(\theta)+f(-\theta))/2$, we will show that $h(\gamma)$ is in the convex hull of $h(\alpha)$, $h(\beta)$, $h(\pi-\alpha)$ and $h(\pi - \beta)$.
Explicitly, $h(\theta) = (\cos \theta, 0, \cos (3 \theta), 0)$, so all of these points are in $2$ dimensions.
The points $h(\alpha)$, $h(\beta)$, $h(\pi-\alpha)$ and $h(\pi - \beta)$ are the vertices of a parallelogram with center at $(0,0)$.
As $\gamma$ approaches $\pi/2$, the point $h(\gamma)$ approaches $(0,0)$ so, for $\gamma$ close enough to $\pi/2$, the point $h(\gamma)$ will be inside this parallelogram. $\square$
