Do $A$ and $A^{2}$ share eigenvectors if both are real and symmetric? If $A$ and $A^{2}$ is a real symmetric matrix and $\overrightarrow{\lambda}$ is an eigenvector of $A^{2}$ with eigenvalue $\lambda$ then does that imply that $\overrightarrow{\lambda}$ is an eigenvector of $A$ and furthermore the corresponding eigenvalue is $\pm\sqrt{\lambda}$ ? 
I know this would work if $A^{2}$ is nilpotent but not sure for when that is not the case.

No, not necessarily.  For instance, suppose $A=\begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$.  Then $A^2=I$, so every vector is an eigenvector of $A^2$.  But, for instance, $(1,1)$ is not an eigenvector of $A$.
More generally, if there is a number $c\neq0$ such that both $c$ and $-c$ are eigenvalues of $A$ (with eigenvectors $v$ and $w$, say), then any linear combination of $v$ and $w$ will be an eigenvector of $A^2$ (with eigenvalue $c^2$), but a linear combination $av+bw$ will only be an eigenvalue of $A$ if $a=0$ or $b=0$.
