Let $T:V \rightarrow V$ be given by T(w,z) = (z,w)
We can have $T(w,z) = \lambda(w,z)$
Two conditions: $$z = \lambda w$$ $$w = \lambda z$$ We sub $z$ into $w$ we get $$w = \lambda^2w$$
Then we solve for $\lambda$ we get $$\lambda = \pm 1$$
Eigenvector to eigenvalue:
$\lambda$ = 1 are $$(w,w) = span(1,1)$$ $\lambda = -1 $ are $$(w,-w) = span(1,-1)$$
What is the reason that the following equation end up with $(w,w)$ and $(w,-w)$?
Here I assume that $V$ is $\mathbb K^2$.
These are indeed the two eigenvalues of $T$. Since $T$ is the "swapping coordinates map", we can also see that these are the eigenvectors for the following reasons:
$(1)$ It is clear that, if the two coordinates are equal, then swapping the coordinates does not change the vector. That is, $\lambda=1$ is an eigenvalue since swapping the coordinates of a vector of the form $(w,w)$ does not change the vector.
$(2)$ If a vector has coordinates that only differ by a sign, then swapping the coordinates still yields a vector whose coordinates only differ by a sign. That is, $\lambda=-1$ is an eigenvalue since swapping the coordinates of a vector of the form $(w,-w)$ yields a vector of the form $(-w,w)=-(w,-w)$.
More algebraically, to find the eigenvectors we just solve for the solution systems to $$\begin{bmatrix}\lambda & -1 \\ -1 & \lambda\end{bmatrix}\begin{bmatrix}w\\z\end{bmatrix}=0$$ for $\lambda=\pm 1$. This is because the matrix on the left is $\lambda I-A$ where $A$ is the matrix of $T$.