About your proof. The "only if" ($\Rightarrow$) direction is quite simple. The characteristic polynomial is (as you say) the product of the invariant factors, and the minimal polynomial is similarly the largest invariant factor; both then are determined by the rational canonical form (RCF). So your work for the minimal polynomial is not really needed; moreover it seems to go off in an completely wrong direction (what is the point of focussing on a single root $\lambda$ in $F$ of the characteristic polynomial, which you don't even know exists?), and I cannot follow much of it.
Your proof of the "if" direction is entirely wrong. The characteristic polynomial is certainly not the same as the RCF (they aren't even the same kind of thing), and matrices with the same characteristic polynomials can very well not be similar (for instance one could be diagonalisable but the other not), ever for the $3\times3$ case.
What you do know is that the characteristic polynomial is the product of the invariant factors, and the minimal polynomial is largest of them (which all the others divide). From this one cannot in general conclude that all invariant factors must correspond (so that the matrices will be similar), but now the $3\times3$ condition comes to the rescue. This means the characteristic polynomial$~\chi$ has degree$~3$, and this makes it impossible to obtain equal minimal polynomials yet distinct invariant factors. To be precise, if the common minimal polynomial$~\mu$ has degree at least$~2$, then any remaining invariant factors have as product the quotient $\chi/\mu$ which has degree $1$ at most; either (if $\deg \chi/\mu=0$) there are no other invariant factors, or (if $\deg \chi/\mu=1$) there is one remaining invariant equal to $\chi/\mu$. But this leaves only the case $\deg\mu=1$, in which case the remaining invariant factors must all be monic polynomials dividing$~\mu$, which forces them to be equal to$~\mu$.
A similar argument for the $4\times4$ or larger cases fails. For $n=4$, one could have $\deg\mu=2$ and then the remaining factor $\chi/\mu$ of degree $2$ could either itself be$~\mu$, or could split into two identical invariant factors of degree$~1$. For such a minimal counterexample one needs $\chi$ to be of the form $(X-\lambda)^4$ and $\mu=(X-\lambda)^2$; then the remaining invariant factors are either $(X-\lambda)^2$ or twice $X-\lambda$.
You can see here an example where this happens (with $\lambda=0$).
From the characteristic polynomial, the sum of sizes of the Jordan blocks for eigenvalue $1$ is $5$, the sum of sizes of Jordan blocks for eigenvalue $-3$ is $1$, and those are all the eigenvalues. Thus the matrix is $6 \times 6$, and its Jordan form has a single $-3$ on the diagonal (whether you put this at top left or bottom right, or even between two blocks for eigenvalue $1$, is up to you). From the minimal polynomial, the largest Jordan block for eigenvalue $1$ has size $2$. Thus you
could have either two such blocks of size $2$ and one of size $1$, or one of size $2$ and three of size $1$. Again, the arrangement of the blocks does not matter. Matrices with different arrangements are similar.
As for your second question, "complex" does not mean "not real". It just means that the entries of the matrix are all complex numbers. Your matrices in Jordan form are already "complex". You don't need to put any $i$'s in them.
Best Answer
Working over $\mathbb{F}_2$ makes this pretty simple. If your matrices are invertible, then $1$ is the only eigenvalue they can have. Any two invertible $n\times n$ matrices in $\mathbb{F}_2$ will have characteristic polynomial $p(x)=(x-1)^n$. By choosing the largest Jordan block to be the same size between the two matrices, we get two matrices with equal minimal polynomial $q(x) = (x-1)^k$, where $k$ is the size of the largest block. Finally, for the two to have eigenspaces of equal dimension, we must have an equal number of Jordan blocks (recall that the number of Jordan blocks corresponding to an eigenvalue $\lambda$ is equal to the geometric multiplicity of $\lambda$).
You can check that the largest Jordan block must be larger than $2\times 2$, or you cannot possibly fulfill the above conditions. Thus at least a $3\times 3$ block is needed. It is also sufficient, since we can have $$\begin{pmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix},\ \begin{pmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1\end{pmatrix},$$ which are two non-similar invertible matrices with characteristic polynomial $p(x) = (x-1)^7$, minimal polynomial $q(x) = (x-1)^3$ and eigenspaces of dimension $3$.