The exchange operator is not a operator by itself, it is only defined with the orbital it is working on:
$%
\newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle}
\newcommand{\lb}{\left|}\newcommand{\rb}{\right|}
\newcommand{\op}[1]{\mathbf{#1}}$
$$\begin{align}
&& \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\
\text{with}&&
\op{J}_j\lb \phi_i\rr &=
\ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
\lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\
\text{and}&&
\op{K}_j\lb \phi_i\rr &=
\ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
\lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr.
\end{align}$$
While in the Coulomb operator case the $\lb \phi_i\rr$ doesn't do anything, so you can take it through the proof as a constant:*
\begin{align}
\sum_j\op{J}_j\lb \phi_i\rr
&= \sum_j
\ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
\lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
&= \sum_k \sum_l \sum_j U_{kj}^*U_{lj}
\ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
\lb \phi_l(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
&= \sum_k \sum_l \delta_{kl}
\ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
\lb \phi_l(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
&= \sum_k
\ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
\lb \phi_k(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr\\
&= \sum_k \op{J}_k \lb \phi_i\rr
\end{align}
That is not the case in the Exchange operator, as it switches the orbitals:
\begin{align}
\sum_j\op{K}_j\lb \phi_i\rr
&= \sum_j
\ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
\lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr\\
&= \sum_k \sum_l \sum_j U_{kj}^*U_{lj}
\ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
\lb \phi_i(\op{x}_1) \rr \lb \phi_l(\op{x}_2) \rr\\
&= \sum_k \sum_l \delta_{kl}
\ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
\lb \phi_i(\op{x}_1) \rr \lb \phi_l(\op{x}_2) \rr\\
&= \sum_k
\ll \phi_k(\op{x}_1) \rb r_{12}^{-1}
\lb \phi_i(\op{x}_1) \rr \lb \phi_k(\op{x}_2) \rr\\
&= \sum_k \op{K}_k \lb \phi_i\rr
\end{align}
* Let
$$
\mathbb{U}=\left(\begin{matrix}
U_{11} & U_{12} & \dots \\
U_{21} & U_{22} & \dots \\
\vdots & \vdots & \ddots\\
\end{matrix}\right);
\mathbb{U}^\dagger\mathbb{U}=\mathbb{E}
\implies U_{ij}^* U_{kl}= \delta_{ij} = \begin{cases}
1; & i = j = k = l \\ 0; & i \neq j \dots\\ \end{cases}
.$$