I am trying to solve the following extension problem: find all exact sequences of the form: $$ 0\longrightarrow \mathbb{F}_p^n\longrightarrow G \longrightarrow C_p\longrightarrow 0$$
Since $\mathbb{F}_p^n$ is Abelian, it will be a matter of computing $H^2(C_p, \mathbb{F}_p^n)$ for each possible $C_p$-action on $\mathbb{F}_p^n$. All such actions correspond to order-$p$ elements of GL$_n(\mathbb{F}_p)$. These matrices satisfy $(X-1)^p=X^p-1=0$, ande therefore have a Jordan Normal form with ones on the diagonal (and where the Jordan blocks have size $\leqslant n$). Since $H^2(C_p,-)$ preserves products, it therefore suffices to assume $n\leqslant p$, and that the matrix of the $C_p$-action is given by:
$\begin{bmatrix} 1 & 1 & 0 & \dots & 0 & 0 \\ 0 & 1 & 1 & \dots & 0 & 0 \\ \dots & \dots & \dots &\dots&\dots &\dots \\ 0 & 0 & 0 & \dots & 1 & 1\\ 0 & 0 & 0 &\dots&0&1 \end{bmatrix}$
To compute $H^2(C_p,\mathbb{F}_p^n)$, I first observed that there is an exact sequence of $C_p$-groups:
$$ 0\longrightarrow \mathbb{F}_p^{n-1}\longrightarrow \mathbb{F}_p^n \longrightarrow \mathbb{F}_p\longrightarrow 0$$
The group $C_p$ acts trivially on $\mathbb{F}_p$, and in an identical-as-before fashion on $\mathbb{F}_p^{n-1}$. The question therefore becomes a recursive question. We gain the following L.E.S:
$$ 0\longrightarrow (\mathbb{F}_p^{n-1})^{C_p}\longrightarrow (\mathbb{F}_p^n)^{C_p} \longrightarrow (\mathbb{F}_p)^{C_p}\longrightarrow H^1(C_p,\mathbb{F}_p^{n-1})\longrightarrow H^1(C_p,\mathbb{F}_p^n)\longrightarrow\dots\\\dots\longrightarrow H^1(C_p,\mathbb{F}_p)\longrightarrow H^2(C_p,\mathbb{F}_p^{n-1})\longrightarrow H^2(C_p,\mathbb{F}_p^n)\longrightarrow H^2(C_p,\mathbb{F}_p)$$ It is easy to see that $(\mathbb{F}_p^{n-1})^{C_p}=(\mathbb{F}_p^n)^{C_p}=(\mathbb{F}_p^n)^{C_p}=\mathbb{F}_p$. I also found that $H^1(C_p,\mathbb{F}_p^n)=\mathbb{F}_p$ if $n<p$, and that $H^1(C_p,\mathbb{F}_p^n)=0$ if $n=p$. I also found that $H^1(C_p, \mathbb{F}_p)=H^2(C_p,\mathbb{F}_p)=\mathbb{F}_p$. Many arrows of the exact sequence become isos or zero maps, so the exact sequence becomes much shorter. If $n<p$, we get:
$$ 0\longrightarrow H^2(C_p,\mathbb{F}_p^{n-1})\longrightarrow H^2(C_p,\mathbb{F}_p^n)\longrightarrow \mathbb{F}_p$$ If $n=p$, we get: $$ 0\longrightarrow\mathbb{F}_p\longrightarrow H^2(C_p,\mathbb{F}_p^{n-1})\longrightarrow H^2(C_p,\mathbb{F}_p^n)\longrightarrow \mathbb{F}_p$$
The problem I am dealing with, is what's next. Or more precisely, how should I (dis)prove that the map $H^2(C_p,\mathbb{F}_p)\to\mathbb{F}_p=H^2(C_p,\mathbb{F}_p)$ is surjective. I know a generating cocycle for $H^2(C_p,\mathbb{F})$, but how can I see it is in the image of the map?
Let $M$ be a $C_p$-module. Let $g$ be a generator of $C_p$, and define maps \begin{gather} \phi:m\in M\mapsto gm-m\in M, \\ \psi:m\in M\mapsto m+gm+\cdots+g^{p-1}m\in M. \end{gather} Then one can show that $$H^2(C_p,M)\cong\frac{\ker\phi}{\operatorname{im}\psi}.$$ In your specific case, you have $M=F_p^n$ with $n\leq p$ and $g$ acting through the matrix $J_n$ which is a Jordan block with eigenvalue $1$. The kernel of $\phi$ is just the subspace of $g$-invariant elements of $M$, that is, the eigenspace of $g$ for the eigenvalue $1$, and it clearly is one-dimensional, spanned by $e_1$, the first vector in the standard basis of $M$. It follows from this that $H^2$ is at most one-dimensional. And it is zero exactly when $\psi\neq0$