Direct proof that Pr[2 immediately follows 1] in a random permutation is 1/n The probability that $1$ is a fixed point of a random permutation of $\{1,2,\ldots,n\}$ (with uniform distribution) is $1/n.$  This is easy to prove since there are $(n-1)!$ permutations that have $1$ as a fixed point and $(n-1)!\,/\,n!=1/n.$  A more direct proof is simply to observe that the image of $1$ under a random permutation is equally likely to be any of the $n$ elements of $\{1,2,\ldots,n\}.$
The probability that $2$ immediately follows $1$ in a random permutation is also $1/n,$ and there is a proof similar to the first proof above in that it involves the computation $(n-1)!\,/\,n!=1/n.$  In a permutation $\pi_1\pi_2\ldots\pi_n$ such that $\pi_i\pi_{i+1}=12$ there are $n-1$ possible values of $i$ and there are $(n-2)!$ ways to permute $\{3,4,\ldots,n\}$ among the remaining $n-2$ positions.  Hence there are $(n-1)!$ permutations in which $2$ immediately follows $1.$  Perhaps simpler is to observe that there are $(n-1)!$ permutations of $\{12,3,4,\ldots,n\},$ where $12$ is regarded as a single object.
My question: is there a direct way of seeing this, similar to the direct way of seeing that the probability that $1$ is a fixed point is $1/n?$
 A: You need to ensure the last element in the $n$-element permutation is not $1$, since that would mean it has no immediate successor. The chance of this occurring is $(n-1)/n$.
In a scenario for which $1$ appears among the first $n-1$ entries, there are still $n-1$ elements left that could follow it, viz., $2, \ldots, n$. The probability in a given scenario that $2$ is the immediate successor, then, is $1/(n-1)$.
Then the probability in question is $\frac{n-1}{n} \cdot \frac{1}{n-1} = \frac{1}{n}$ as desired. QED
A: $1$ is equally likely to be in any of the $n$ positions in the permutation.
If it is in the first $n-1$ positions (total probability $\frac{n-1}{n}$), 
then it is equally likely to be followed by any
of the other $n-1$ symbols, and so the probability that it is followed by $2$ is
$\frac{n-1}{n}\times \frac{1}{n-1} = \frac{1}{n}$. If $1$ is in the $n$-th position,
it cannot be followed by anything. So,
$$P\{1~\text{is followed by}~2\} = \frac{1}{n}.$$
A: An improved version of the observation in the comments to Benjamin Dickman's answer:
There is a $1$-to-$n$ map from the set of permutations of $\{2,3,\ldots,n\}$ to the set of permutations of $\{1,2,\ldots,n\}$ obtained by inserting $1$ in any of $n$ possible positions.  Taking $n=4,$ we have, for example,
$$
243\mapsto\{1243,2143,2413,2431\}.
$$
The probability that the insertion point is immediately in front of $2$ is $1/n.$
