I don't know whether this counts as picturing the bijection, but let's write out all the admissible seatings for a four-passenger plane. We will represent each admissible seating by the cycle of displaced passengers, numbered according to their position in line, with passenger $1$ counting as displaced whether they end up sitting in their own seat or not. The cycle $(1,a,b,c,d)$, for example, will mean $1$ sits in $a$'s seat, $a$ sits in $b$'s seat, $b$ sits in $c$'s seat, $c$ sits in $d$'s seat, and $d$ sits in $1$'s seat. It is necessary that $1<a<b<c<d$.
$$
\begin{array}{r|r||r|r||r}
\text{cycle} & \text{seating} & \text{cycle} & \text{seating} & \text{probability}\\
(1) & 1234 & (1,4) & 4231 & \frac{1}{4}\cdot1\cdot1\cdot1=\frac{1}{4}\\
(1,2) & 2134 & (1,2,4) & 4132 & \frac{1}{4}\cdot\frac{1}{3}\cdot1\cdot1=\frac{1}{12}\\
(1,3) & 3214 & (1,3,4) & 4213 & \frac{1}{4}\cdot1\cdot\frac{1}{2}\cdot1=\frac{1}{8}\\
(1,2,3) & 3124 & (1,2,3,4) & 4123 & \frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot1=\frac{1}{24}\\
\hline
& & & & \frac{1}{2}
\end{array}
$$
In the first row of this table, the cycle $(1)$ represents the situation where passenger $1$ sits in her own seat, while the cycle $(1,4)$ represents the situation where passenger $1$ sits in passenger $4$'s seat and passenger $4$ sits in passenger $1$'s seat. Each of these situations occurs with probability $\frac{1}{4}$. Similarly in row $2$, the cycle $(1,2)$ represents the situation where $1$ sits in $2$'s seat and $2$ sits in $1$'s seat, while the cycle $(1,2,4)$ represents the situation where $1$ sits in $2$'s seat, $2$ sits in $4$'s seat, and $4$ sits in $1$'s seat. Each of these situations occurs with probability $\frac{1}{12}$. The remaining two rows can be understood similarly. Observe that the total probability of the situations represented by the first column and the total probability of the situations represented by the third column both equal $\frac{1}{2}$. In column $1$, passenger $4$ sits in his own seat; in column $3$, passenger $4$ sits in passenger $1$'s seat.

The admissible seatings for a five-passenger plane are below.
$$
\begin{array}{r|r||r|r||r}
\text{cycle} & \text{seating} & \text{cycle} & \text{seating} & \text{probability}\\
(1) & 12345 & (1,5) & 52341 & \frac{1}{5}\cdot1\cdot1\cdot1\cdot1=\frac{1}{5}\\
(1,2) & 21345 & (1,2,5) & 51342 & \frac{1}{5}\cdot\frac{1}{4}\cdot1\cdot1\cdot1=\frac{1}{20}\\
(1,3) & 32145 & (1,3,5) & 52143 & \frac{1}{5}\cdot1\cdot\frac{1}{3}\cdot1\cdot1=\frac{1}{15}\\
(1,4) & 42315 & (1,4,5) & 52314 & \frac{1}{5}\cdot1\cdot1\cdot\frac{1}{2}\cdot1=\frac{1}{10}\\
(1,2,3) & 31245 & (1,2,3,5) & 51243 & \frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\cdot1\cdot1=\frac{1}{60}\\
(1,2,4) & 41325 & (1,2,4,5) & 51324 & \frac{1}{5}\cdot\frac{1}{4}\cdot1\cdot\frac{1}{2}\cdot1=\frac{1}{40}\\
(1,3,4) & 42135 & (1,3,4,5) & 52134 & \frac{1}{5}\cdot1\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot1=\frac{1}{30}\\
(1,2,3,4) & 41235 & (1,2,3,4,5) & 51234 & \frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot1=\frac{1}{120}\\
\hline
& & & & \frac{1}{2}
\end{array}
$$