Find an affine linear map $$\mathbb{Z}_2^5\to\mathbb{Z}_2^5$$ that sends $(0,1,0,0,1)$ to $(1,0,0,1,0)$.
So I know that an affine linear map is one of the form $Az+b$ where $b,z\in\mathbb{Z}_2^5$ and $A\in\mathbb{Z}_2^{5\times5}$. So we want to solve the congruence
$$\begin{pmatrix}a_{11}&a_{12}&a_{13}&a_{14}&a_{15}\\a_{21}&a_{22}&a_{23}&a_{24}&a_{25}\\a_{31}&a_{32}&a_{33}&a_{34}&a_{35}\\a_{41}&a_{42}&a_{43}&a_{44}&a_{45}\\a_{51}&a_{52}&a_{53}&a_{54}&a_{55}\end{pmatrix}\begin{pmatrix}0\\1\\0\\0\\1\end{pmatrix}+\begin{pmatrix}b_1\\b_2\\b_3\\b_4\\b_5\end{pmatrix}\equiv\begin{pmatrix}1\\0\\0\\1\\0\end{pmatrix}\pmod{2}$$
That is,
\begin{align} a_{12}+a_{15}+b_1 &\equiv 1\pmod{2}\\ a_{22}+a_{25}+b_2 &\equiv 0\pmod{2}\\ a_{32}+a_{35}+b_3 &\equiv 0\pmod{2}\\ a_{42}+a_{45}+b_4 &\equiv 1\pmod{2}\\ a_{52}+a_{55}+b_5 &\equiv 0\pmod{2} \end{align}
How would one solve this? Is there an easier way to solve this?
Choose arbitrary $A$, and then $$b = \pmatrix{1\cr 0\cr 0 \cr 1 \cr 0\cr} - A \pmatrix{0\cr 1\cr 0\cr 0\cr 1\cr} \ \mod 2$$