We claim that the answer is no. Note that $n$ must be odd. We first write the polynomial $P(x)$ in the form $a_{2m+1}x^{2m+1}+a_{2m-1}x^{2m-1}+\cdots+a_1x$, where we do not have any terms of even degree as $P(x)$ is odd (i.e. $P(-x)=-P(x)$ for all $x$). Its derivative is $Q(x)=a_{2m+1}(2m+1)x^{2m}+a_{2m-1}(2m-1)x^{2m-2}+\cdots+a_1$. The key claim is the following: Claim: Suppose $(a,P(a))$ and $(b,P(b))$ are consecutive points in the sequence. Then $a$ and $b$ must have different signs. Proof: The tangent passing through $(a,P(a))$ has equation $y=Q(a)(x-a)+P(a)$. Substituting in $x=b$, we have $$P(b)=Q(a)(b-a)+P(a)$$Re-writing, we have $$Q(a)=\frac{P(b)-P(a)}{b-a}=a_{2m+1}\cdot\frac{b^{2m+1}-a^{2m+1}}{b-a}+a_{2m-1}\cdot\frac{b^{2m-1}-a^{2m-1}}{b-a}+\cdots+a_1$$We group the terms of degree $2m$ together to get \[((2m+1)a^{2m}-b^{2m}-b^{2m-1}a-\dots-a^{2m})a_{2m+1}\] Suppose $a$ and $b$ have the same sign. If $|a|>|b|$, then the expression in the brackets is positive, and if $|a|<|b|$, then the expression is negative. Similarly, if we group the terms of degrees $2m-2, 2m-4, \cdots,0$ together, we obtain a consistent sign for all the expressions, which means that the equation $$Q(a)-\frac{P(b)-P(a)}{b-a}=0$$has no solutions, contradiction. The conclusion follows immediately - since $n$ is odd, this means that $A_n$ and $A_1$ have to lie on the same side of the $y$-axis, contradiction.

Another way of proving that consequtive points are on different sides of $oY$ Suppose $(a,P(a))$ and $(b,P(b))$ are consecutive points in the sequence. Then $\exists c \in (a,b) : P'(c) = \dfrac{P(b)-P(a)}{b-a} = P'(a)$. Therefore $P'(x) - P'(a)$ has two different roots in $[a,b)$. Therefore $P''(x)$ has a root in $(a,b)$, but since $P''(x) + P''(-x)=0$ and $P''(x)$ has only nonnegative coefficients this root must be equal to $0$. Therefore $0 \in (a,b)$ which is the same as saying $a$ and $b$ have different signs.