Solving $BA=\lambda A$ where $A$ and $B$ are $2$ by $2$ matrices . $BA=\lambda A$ where $A$ and $B$ are $2$ by $2$ matrices. 
and $$B= \quad \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \quad$$
 A: $B$ is a permutation matrix: when $A$ is multiplied by $B$ on the left, it swaps its rows (its columns if multiplied on the right).
Therefore, if the result is $\lambda A$, then, if $r_1$ and $r_2$ are the rows of $A$:
$$r_1=\lambda r_2$$
$$r_2=\lambda r_1$$
Hence, if the rows are not both zero (i.e., if $A$ is not the zero matrix), $\lambda=\pm1$. And the rows of $A$ are equal up to sign. It's easy to check this is a sufficient condition for the equality to hold.

Eigenvalue approach
As has already been said in the comments above, the columns of $A$ are both eigenvectors of $B$ for the same eigenvalue (or possibly zero). This is so because, if $c_1$ and $c_2$ are the columns of $A$, then $BA=\lambda A$ implies that:
$$Bc_1=\lambda c_1$$
$$Bc_2=\lambda c_2$$
With the same $\lambda$.
The characteristic polynomial of $B$ is $\lambda^2-1$, hence the eigenvalues of $B$ are $\pm 1$.

In the case of eigenvalue $\lambda=1$, you have to solve the system
$$\left(\begin{matrix}
-1 & 1\\
1 & -1\end{matrix}\right)
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$
And you find that the eigenvector $(x\; y)^T$ has to verify $x=y$. Henceforth, the rows of $A$ are equal.

In the case of eigenvalue $\lambda=1$, you have to solve the system
$$\left(\begin{matrix}
1 & 1\\
1 & 1\end{matrix}\right)
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$
And you find that the eigenvector $(x\; y)^T$ has to verify $x=-y$. Henceforth, the rows of $A$ are opposite.
If only one of $c_1$ or $c_2$ is zero, that does not change the conclusion, and if both are zero, any $\lambda$ will do.
A: You can just set up the system and solve it. Letting $A_{11}, A_{12}, A_{21},A_{22}$ denote the components of $A$, we get 
$$A_{21} = \lambda A_{11}$$
$$A_{22} = \lambda A_{12}$$
$$A_{11} = \lambda A_{21}$$
$$A_{12} = \lambda A_{22}$$
The first and third equations give us $A_{21} = \lambda^2 A_{21} \implies A_{21} (1-\lambda^2)$. Similarly, the second and fourth equations give us $A_{22}(1 - \lambda^2)=0$. If $\lambda \neq \pm 1$, then $A=0$. Nontrivial solutions are then given by $\lambda = \pm 1$. To get the solutions for $A$, solve the above (now linear) system of equations for each $\lambda$. 
