I have difficulty following the proof that transvections are conjugates in $GL(V)$, and for $n \ge 3$ even in $SL(V)$. I give the necessary definitions and the proof, with the problematic parts marked, hoping for someone to explain them to me.
Let $F$ be a finite field and $V$ a vector space over $K$. Denote by $GL(V)$ the set of all bijective linear transformations on $V$ and by $SL(V)$ the special linear group $SL(V) = \{ \varphi \in GL(V) : \det(\varphi) = 1 \}$. Further for $\varphi \in GL(V)$ and $v \in V$ define $C_V(\varphi) := \{ v \in V : \varphi(v) = v \}$ and $[v, \varphi] = \varphi(v) - v$. For a subspace $W \le V$ denote by $[W, \varphi]$ the subspace generated by the elements $[v,\varphi]$ with $v \in W$.
An element $\alpha \in GL(V)$ is called transvection if
-$[V, \alpha]$ is one-dimensional
-$C_V(\alpha)$ has dimension $n-1$
-$[V,\alpha]$ is a subspace of $C_V(\alpha)$.
Denote by $n$ the dimension of $V$.
Lemma: Let $\alpha, \beta \in GL(V)$ be transvections. Then $\alpha, \beta \in GL(V)$ are conjugate in $GL(V)$. Further if $n \ge 3$ then they are even conjugates in $SL(V)$.
Proof: Let $v, w \in V$ such that $[V, \alpha] = \langle v \rangle$ and $[V,\beta] = \langle w \rangle$.
- Case: $\langle v \rangle = \langle w \rangle$. Then let $\lambda \in F$ be such that $w = \lambda \cdot w$. Is $C_V(\alpha) = C_V(\beta)$ then we have $\alpha = \beta$ (here is the first point I am unsure about, why does equality of two $n-1$ dimensional spaces implies equality of the maps $\alpha$ and $\beta$?) and so they are conjugates. So let $C_V(\alpha) \ne C_V(\beta)$. Then set $W := C_V(\alpha) \cap C_V(\beta)$, this is a $(n-2)$ dimensional subspace (why has it to be $n-2$ dimensional?), in particular $v,w \in W$.
Let $u_1, u_2 \in V$ such that $\{ u_1, u_2 \}$ and a basis of $W$ is a basis of $V$ and such that (why is such a choice possible?) $$ \alpha(u_1) = u_1 + v, \quad \beta(u_1) = u_1, \quad \alpha(u_2) = u_2, \quad \beta(u_2) = u_2 + v. $$ Further let $\tau \in GL(V)$ be such that $\tau(u_1) = u_2, \tau(u_2) = u_1$ and $W \le C_V(\tau)$. [then some calculations follow, which I think I have understood]
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- Case: $\langle v \rangle \ne \langle w \rangle$ [guess this case I have understood]
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Now let $n \ge 3$. We know that $\alpha, \beta \in SL(V)$. Consider $V$ as $K^n$, and choose some ordered standard base $b_1, \ldots, b_n$ and the following set $D \subseteq GL(V)$
$$ D = \{ \gamma \in GL(V) : \mbox{there exists some $\lambda \in K\setminus\{0\}$ such that } \gamma(b_1) = b_1, \gamma(b_2) = \lambda \cdot b_2, \gamma(b_3) = b_3, \ldots, \gamma(b_n) = b_n \} $$ It is $|D| = |K|-1$, and every $\delta \in GL(V)$ could be written as $\delta = \delta_1 \circ \delta_2$ with $\delta_1 \in SL(V)$ and $\delta_2 \in D$, i.e. $GL(V) = SL(V) \circ D$. From the form of the corresponding matrices we see that transvections and elements form $D$ commute, so $\alpha$ and $\beta$ are already conjugates in $SL(V)$. QED
The last part for $n \ge 3$ is also nebulous to me, where is $n \ge 3$ exactly used?
First off, as a note, if $F$ is algebraically closed, I think this question is best done using the Jordan Normal Form.
Let me first do some general calculations. Let $\alpha$ be a transvection. Suppose $\{v_{1}, \ldots, v_{n-1}\}$ be a basis for $C_{V}(\alpha)$ and let $u$ be a complementary vector. Let $u'= \alpha(u) - u$. Then, $u' = cu + v'$ for some $v' \in C_{V}(\alpha)$. Applying $\alpha$ to $u'$, we get
$$\alpha(u') - u' = c^{2}u + cv' + v' - cu - v' = (c^{2} - c)u + cv'$$ which must be a scalar multiple of $u'$. This implies $c^{2} - c = c^{2}$ and hence $c = 0$.
So, in general $u' = \alpha(u) - u \in C_{V}(\alpha)$.
(In terms of the Jordan Form, this means that our matrix has one Jordan block, with 1's on the diagonal and exactly one off diagonal term. This is actually enough to get conjugacy in $GL(V)$ if $F$ is algebraically closed.)
Now to your questions. Under the hypotheses of your first question, we have $C_{V}(\alpha) = C_{V}(\beta)$. So taking $\{v_{1}, \ldots, v_{n}\}$ as a basis for this subspace and $u$, a complementary vector, we already have $$\alpha(v_{i}) = \beta(v_{i}) = v_{i}$$ so we need $\alpha(u) = \beta(u)$ to get $\alpha = \beta$.
By our previous calculations, we have $\alpha(u) = u + v'$, $\beta(u) = u + v''$ for $v', v'' \in C_{V}(\alpha)$. The equality of $\langle v\rangle$ and $\langle w \rangle$ tell us that $v' = \lambda v''$ but I do not think they have to be equal. So, $\alpha$ does not have to be equal to $\beta$. However, we can show that they are conjugate:
Suppose $\alpha(u) = u + v'$ and $\beta(u) = u + \lambda v'$. Then, the matrix of $\alpha$ in the basis $\{v_{1}, \ldots, v_{n-1}, u\}$ and the matrix of $\beta$ in the basis $\{\lambda v_{1}, \ldots, \lambda v_{n-1}, u\}$ are the same. Hence, $\alpha$ and $\beta$ are conjugate.
Your second question is easier: If $C_{V}(\alpha) \not = C_{V}(\beta)$, then these are distinct $n-1$ dimensional subspaces. So, we have $C_{V}(\alpha) + C_{V}(\beta) = V$ and hence by the formula $$\dim(U) + \dim(U') = \dim(U + U') - \dim(U \cap U')$$ we must have that $\dim(C_{V}(\alpha) \cap C_{V}(\beta)) = n-2.$
For your third question, take a basis $\{v_{1}, \ldots, v_{n}\}$ such that the first $n-2$ form a basis for $W$, the first $n-1$ form a basis for $C_{V}(\alpha)$, which necessarily implies that $v_{n-1} \notin C_{V}(\beta)$ and hence we can choose the last vector such that $\{v_{1}, \ldots, v_{n-2}, v_{n}\}$ is a basis for $C_{V}(\beta).$ Then, after multiplying by scalars if necessary,
$$\alpha(v_{n-2}) -v_{n-2} = v, \beta(v_{n-1}) - v_{n-1} = w,$$
and we can again multiply $v_{n-1}$ by a scalar to get $\beta(v_{n-1}) - v_{n-1} = v.$
I am not sure about your last question sorry.