I'm having trouble with the Concept of finite fields. Say $\mathbb{F}_2 = \{0,1\}$. How many linear maps from $\mathbb{F}_2^5 \to \mathbb{F}_2^5$ (mapped from five tuples to five tuples) are injective? I know that the null has to be zero, but I'm thinking there might be an infinite number of maps.
I also had another follow up question. So now suppose $U = \{(x,x,y,y,y)\}$ where $x,y\in \mathbb{F}_2$. Using the same mapping, how many linear maps have their null space contain $U$ and equal $U$?
Each linear map $A: \mathbb{F}_2^5 \to \mathbb{F}_2^5$ corresponds to a unique $5\times 5$ matrix with entries in $\mathbb{F}_2$. There are $2^{25}=33554432$ such matrices. This immediately rules out the possibility that there are an infinite number of maps $A: \mathbb{F}_2^5 \to \mathbb{F}_2^5$ (injective or otherwise).
By the rank-nullity theorem, each injective linear map $A: \mathbb{F}_2^5 \to \mathbb{F}_2^5$ corresponds to a unique matrix in the general linear group $GL(5, \mathbb{F}_2)$. The order of the group is
$$ \begin{align} \prod_{k=0}^{n-1}(2^n - 2^k) &= (2^5 - 1)(2^5 - 2)(2^5 - 4)(2^5 - 8)(2^5 - 16) \\ &=31 \cdot 30 \cdot 28 \cdot 24 \cdot 16 = 9999360 \end{align}\tag1 $$
which means that $\frac{9999360}{2^{25}} \approx 29.8\%$ of the matrices in $\mathbb{F}_2^{5\times 5}$ correspond to injective maps.
The formula $(1)$ captures the fact that the first column of an invertible matrix in $\mathbb{F}_2^{5\times 5}$ can be anything but the zero vector, the second can be anything but the zero vector and the first column, the third anything but any of the four linear combinations of the first two columns and so on.