= Solution
For ordered bases $\mathcal B=(v_1,\ldots,v_n)$ of $V$ and $\mathcal C=(w_1,\ldots,w_m)$ of $W$, the <matrix representation of a linear map> $\gamma:V\to W$ is the matrix whose $j$th column consists of the $\mathcal C$-coordinates of $\gamma(v_j)$. Thus
$$
[\gamma(v)]_{\mathcal C}
=[\gamma]_{\mathcal C\leftarrow\mathcal B}[v]_{\mathcal B}.
$$
The operators $\alpha,\beta:V\to V$ are <conjugate linear operators> when there is an isomorphism $s:V\to V$ such that
$$
\beta=s^{-1}\alpha s.
$$
In one basis their matrices therefore satisfy $[\beta]=[s]^{-1}[\alpha][s]$, so they are similar matrices.
For invertible $\beta$, the map
$$
\phi_\beta(A)=\beta^{-1}A\beta
$$
is linear, and its inverse is $A\mapsto\beta A\beta^{-1}$; hence it is a linear isomorphism of $\operatorname{End}(V)$. If $\beta'=s^{-1}\beta s$, define $\Phi_s(A)=s^{-1}As$. A direct substitution gives
$$
\phi_{\beta'}=\Phi_s\phi_\beta\Phi_s^{-1},
$$
so $\phi_{\beta'}$ and $\phi_\beta$ are conjugate. This is the <conjugation operator on an endomorphism space>.
It remains to compute the <Jordan normal form> over $\mathbb C$. By the preceding conjugacy, we may put $\beta$ in Jordan form.
If
$$
\beta=\begin{pmatrix}\lambda&0\\0&\mu\end{pmatrix},
\qquad \lambda\mu\ne0,
$$
then the matrix units $E_{ij}$ are eigenvectors because
$$
\phi_\beta(E_{ij})=\frac{\beta_j}{\beta_i}E_{ij}.
$$
Thus
$$
\boxed{\operatorname{JNF}(\phi_\beta)
=\operatorname{diag}\left(1,1,\frac\mu\lambda,\frac\lambda\mu\right)}.
$$
This includes the scalar case $\lambda=\mu$, when $\phi_\beta$ is the identity.
Otherwise $\beta$ has one size-two <Jordan block>. Multiplying $\beta$ by a nonzero scalar does not change $\phi_\beta$, and conjugating within its Jordan class allows us to use $J=I+N$ with $N=E_{12}$. Put $D=\phi_J-I$. Direct multiplication gives
$$
D(E_{11})=E_{12},\quad D(E_{12})=0,\quad
D(E_{22})=-E_{12},
$$
$$
D(E_{21})=E_{22}-E_{11}-E_{12}.
$$
Hence $D^3=0$, $\operatorname{rank}D=2$, and $\operatorname{rank}D^2=1$. The nilpotent Jordan blocks of $D$ therefore have sizes three and one. Adding the identity gives
$$
\boxed{\operatorname{JNF}(\phi_\beta)=J_3(1)\oplus J_1(1)}.
$$
These two cases are the <Jordan normal form of conjugation on two-by-two matrices>.
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