Let $X$ be the wedge sum of two $2$dimensional spheres and $f$ a continuous function from $X$ into $X$. Does $f$ have either a fixed point or a periodic point of order 2? Thanks

Will's comment above shows that the action of $f$ on $H^2$ is of order 3 (the eigenvalues are the nontrivial third roots of unity) and something like $\begin{pmatrix}0&1\\1&1\end{pmatrix}$. So I thought one should try a construction using third roots of unity realizing this. Write the wedge as $S^1 \times [0,2] / \sim $ where one identifies $(s,0)\sim (s',0), (s,1)\sim (s',1), (s,2)\sim (s',2)$. Let $\zeta$ be a third root of unity, and define $f: S^1 \times [0,2] / \sim\ \to \ S^1 \times [0,2] / \sim$ by $f(s,t)=(\zeta s, 22t)$ for $t\le 1$ and $f(s,t)=(\zeta s, t1)$ for $t\ge 1$ Note that the three points where $t=0,t=1,t=2$ are permuted cyclically. This implies $f$ and $f^2$ have no fixed points (for all other points the $\zeta$ respectively $\zeta^2$ factor works). Note also that multiplication by $\zeta$ is not really necessary, almost any rotation of the circle would be fine. Multiplication by $\zeta$ allows $f^3$ to be close to the identity. If one uses a different rotation, the three points with $t=0,t=1,t=2$ are the only fixed points. 

