Immediately, we can see that
$$
A = |1\rangle\langle0| + |0\rangle\langle1|.
$$
If the input and out bases are $\{|0\rangle, |1\rangle\}$, then
$$
|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \quad\textrm{and}\quad \langle0| = \begin{pmatrix} 1 & 0 \end{pmatrix}, \quad \langle1| = \begin{pmatrix} 0 & 1 \end{pmatrix},
$$
so we can write the first equation as
$$
\begin{align}
A &= \begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \begin{pmatrix} 0 & 1 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix} \otimes \begin{pmatrix} 1 & 0 \end{pmatrix} \\
&= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\\
&= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}
\end{align}
$$
to solve the first question. Note that in this case, $A$ is the equal to the bitflip or Pauli $X$ operation.
Secondly, for fun, let us choose to write $A$ in input basis $\{|+\rangle, |-\rangle\}$, where
$$
|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \quad\textrm{and}\quad |-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)
$$
and output basis $\{|L\rangle, |R\rangle\}$, where
$$
|L\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle) \quad\textrm{and}\quad |R\rangle = \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle).
$$
Rewriting our original $\{|0\rangle, |1\rangle\}$ bases vectors in terms of the above bases, we find
$$
\begin{align}
|0\rangle &= \frac{1}{\sqrt{2}}(|+\rangle + |-\rangle) = \frac{1}{\sqrt{2}}(|L\rangle + |R\rangle) \quad \textrm{and}\\
|1\rangle &= \frac{1}{\sqrt{2}}(|+\rangle - |-\rangle) = \frac{i}{\sqrt{2}}(|R\rangle - |L\rangle),
\end{align}
$$
with $\langle0| = (|0\rangle)^\dagger$ and $\langle1| = (|1\rangle)^\dagger$.
From this we can rewrite $A$ as
$$
\begin{align}
A &= \frac{i}{2}(|R\rangle - |L\rangle)(\langle+| + \langle-|) + \frac{1}{2}(|L\rangle + |R\rangle)(\langle+| - \langle-|) \\
&= \frac{1}{2}\big[ (1+i)|R\rangle\langle+| + (1-i)|L\rangle\langle+| + (-1+i)|R\rangle\langle-| + (-1-i)|L\rangle\langle-|\big] \\
&= \frac{1+i}{2}(|R\rangle\langle+| - i|L\rangle\langle+| + i|R\rangle\langle-| - |L\rangle\langle-|)
\end{align}
$$
To get $A$ in the desired matrix form, we then set
$$
|L\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad |R\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \quad\textrm{and}\quad \langle+| = \begin{pmatrix} 1 & 0 \end{pmatrix}, \quad \langle-| = \begin{pmatrix} 0 & 1 \end{pmatrix},
$$
such that
$$
A = \frac{1+i}{2} \begin{pmatrix} -i & -1 \\ 1 & i \end{pmatrix}.
$$