Consider a set $M$ of all possible square matrices over a finite field $F_p$. Now consider a map $f_A(x)=A.x$ where $x$ $\in$ $M$ and also the matrix $A$ is a member of $M$. It is needless to mention that the set M forms a vector space over field itself with the module p addition and modulo p scalar multiplication. So the map $f_A$ is linear.
Consider a dynamical system $X_{t+1}=f_A(X_t)$. Suppose there is an orbit of a vector $u$ $\in$ $M$ is $A.u$, $A^2.u$...$A^k.u$.... Let $v=Pu$ where $P$ is another matrix from the set M. So the vector v is generated by a linear operator $P$.
Question is: Is it possible to derive a matrix B which is ''similar'' to the matrix A such that the orbit of the vector $v$ with respect to the map $f_B$ on $M$ is $B.v$, $B^2.v$...$B^k.v...$ ?
You require $Bv = BPu = Au$ and $B^2 v = B^2 Pu = A^2 u$. This requires $B^2Pu = BAu = A^2u$. So $BP$ has all the same eigenvalues as $A$ and also $B^2P$ has all the same eigenvalues as $A^2$. These together imply $B$ has the same eigenvalues as $A$ and $P$ has only the repeated eigenvalue $1$ (and is orthogonal). So your $B$ only exists if $P$ is an orthonormal rotation of the coordinates.