$\def\I{\mathrm{Id}}\def\Mat{\operatorname{Mat}}$Let $J = \lambda\I + N \in \Mat_n(K)$ with \[ N = \begin{pmatrix} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots && \ddots &\\ 0 & \cdots & & 0 & 1\\ 0 & \cdots & & 0 & 0 \end{pmatrix} \] be a Jordan block. Then $J^t$ is just $J$ with the ones below the diagonal. To find a $Q$ as wished, observe how $J$ acts on the unit basis: \[ Je_i = \lambda e_i + Ne_i = \lambda e_i + e_{i-1}, \quad 1 \le i \le n \] (with $e_0 := 0$ for brevity). So $e_1$ is the eigenvector of the Jordan chain $e_1, \ldots, e_n$. We want $Q$ such that $J^t = QJQ^t$ acts on $e_i$ as follows \[ QJQ^te_i = J^t e_i = \lambda e_i + e_{i+1} \] As $Q$ should denote a permutation matrix, write $Qe_i = e_{\pi(i)}$ for some $\pi \in S_n$, then $Q^t e_i = e_{\pi^{-1}(i)}$ for all $i$. We get \[ \lambda e_i + e_{i+1} = QJQ^t e_i = QJe_{\pi^{-1}(i)} = Q\bigl(\lambda e_{\pi^{-1}(i)} + e_{\pi^{-1}(i) - 1}\bigr) = \lambda e_i + e_{\pi(\pi^{-1}(i) - 1)}\] So we want to have $i + 1 = \pi(\pi^{-1}(i) - 1)$ that is $\pi^{-1}(i+1) = \pi^{-1}(i) - 1$, so $\pi^{-1}(i) = \pi(i) = n+1-i$ for each $i$. So \[ Q := \bigl( \delta_{n+1-i,j}\bigr)_{i,j} \] is as wished.
Now let $J$ be a Jordan matrix, \[ J = \begin{pmatrix} J_1 & \cdots & 0 \\ & \ddots\\ 0 & \cdots & J_k \end{pmatrix} \] composed of Jordan blocks $J_l = \lambda_l \I_{n_l} + N_{n_l} \in \Mat_{n_l}(K)$ as above. If $Q_{n_l}$ denotes the $Q$ above of size $n_l$, then \[ Q = \begin{pmatrix} Q_{n_1} & \cdots & 0 \\ & \ddots\\ 0 & \cdots & Q_{n_k} \end{pmatrix} \] is a block diagonal permutation matrix with $QJ Q^t = J^t$.