Say I have a matrix $A$ of $r=rank(A)=1$
I know that in the characteristic polynomial the algebraic multiplicity of $(\lambda-0)$ is $n-r$ which in my case is $n-1$
Is there a rule about the algebraic multiplicity in the minimal polynomial in this case or in the general case?
Every matrix of rank 1 is similar to one of the Jordan forms $$ \left(\begin{array}{cc|ccc}0&1&\\&0&\\ \hline && 0 \\ &&&\ddots \\ &&&&0 \end{array}\right) \qquad\text{or}\qquad \left(\begin{array}{c|ccc}\lambda&\\\hline &0\\&&\ddots\\&&&0\end{array}\right) $$ for some $\lambda\ne 0$.
The first has characteristic polynomial $t^n$ and minimal polynomial $t^2$. The second has caracteristic polynomial $(t-\lambda)t^{n-1}$ and minimal polynomial $(t-\lambda)t$ (or just $t-\lambda$ if $n=1$).
The algebraic multiplicity of $0$, being at least the geometric one, is at least $n-1$ in your case, more generally $n - r$ with $r$ the rank. Contrary to what you say it is not necessarily equal to this value.
Now, it is always at most $n$, the dimension. Thus, in the characteristic polynomial it is either $n-1$ or $n$. Both can happen. For the $n-1$ consider a matrix $0$ except for $a_{11} = 1$. For the latter consider a matrix $0$ except for $a_{12} = 1$. For a rule, you could say it is $n$ if the matrix is nilpotent, and $n-1$ otherwise. Note that the nilpotent matrices are precisely those with $0$ having algebraic multiplicity $n$.
For the minimal polynomial we get that in the former case the algebraic multiplicity of $0$ is $1$. This is because there is another eigenvalue besides $0$ so the dimension of the characteristic space associated to $0$ cannot be the full space and thus has dimension at most $n-1$, that is it coincides with the kernel of $A=A^1$.
In the latter case, one gets that the algebraic multiplicity $m$ in the minimal polynomial is $2$. It cannot be more than two as $\ker A^1 \subset \ker A^2 \dots \subset \ker A^m$ needs to have strict inclusions and since the dimension of $\ker A$ is already $n-1$ there can only be one additional entry in this chain. It also cannot be $1$, as this would mean $\ker A$ is the full space, which is impossible for a rank $1$ matrix.
Matrices of rank$~1$ are easy to understand due to their large eigenspace for $\lambda=0$. Let $A$ be square of size$~n$ and have trace$~c$. Then its characteristic polynomial is $X^{n-1}(X-c)$, so the algebraic multiplicity of the eigenvalue$~0$ is either $n-1$ when $c\neq0$, or $n$ when $c=0$. The minimal polynomial is $X(X-c)$ provided $n>1$ (when $n=1$ the minimal polynomial is of course $X-c$). I explained these things in this answer.
The image of$~A$ is always en $A$-stable subspace, and for a rank$~1$ matrix it has dimension$~1$, so any spanning vector of the image is an eigenvector. The eigenvalue of such a vector equals the trace$~c$ of$~A$, and therefore $X(X-c)$ annihilates $A$. In particular the image of$~A$ is contained in its kernel if and only if $c=0$, and (only) in this case $A$ is not diagonalisable.
