Let $R$ be the ring of $3 \times 3$ matrices with coefficients in $\Bbb Z_5$.
For every $g \in GL_3(\Bbb Z_5)$ prove that the function $$f\colon R \rightarrow R$$ defined as $$x \mapsto g^{-1}xg$$ is an automorphism of $R$.
If I choose the matrix $$g = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix},$$ how many matrices $x\in R$ such that $f(x)=x$ are there?
I have no idea how to solve this. Any help?
This is much more general: suppose $R$ is a ring and $g$ is invertible in $R$. Then the map $f\colon R\to R$, $f(x)=gxg^{-1}$ is an automorphism of $R$.
Indeed, $f(x+y)=g(x+y)g^{-1}=gxg^{-1}+gyg^{-1}=f(x)+f(y)$ and $$ f(xy)=gxyg^{-1}=gxg^{-1}gyg^{-1}=f(x)f(y) $$ Obviously, $f(1)=1$.
Since the map $x\mapsto g^{-1}xg$ is the inverse of $f$, we are done.
For the second part, you want $gxg^{-1}=x$, so $gx=xg$. If $$ x=\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} $$ then $$ gx=\begin{pmatrix} a_{31} & a_{32} & a_{33} \\ a_{21} & a_{22} & a_{23} \\ a_{11} & a_{12} & a_{13} \\ \end{pmatrix} \qquad xg=\begin{pmatrix} a_{13} & a_{12} & a_{11} \\ a_{23} & a_{22} & a_{21} \\ a_{33} & a_{32} & a_{31} \end{pmatrix} $$ Thus you get \begin{cases} a_{31}=a_{13} \\ a_{32}=a_{12} \\ a_{33}=a_{11} \\ a_{21}=a_{23} \end{cases} and so the linear system has rank $4$.