This answer focuses on bit permutations as used in the context of DES for bit permutations IP, IP-1, and E. It numbers bits starting from $1$, as in DES.
A bit permutation is a function $g$ on the set $\{0,1\}^n$ of $n$-bit bistrings. It's entirely defined by a vector of $n$ distinct integers $p_i$ with $1\le p_i\le n$, and the property that for any bitstring $x$ and any integer $i$ with $1\le i\le n$, bit number $i$ of the image of bitstring $x$ by function $g$ is bit number $p_i$ of $x$.
In other word, each input bit goes to it's uniquely assigned output bit, and output bit $i$ is input bit $p_i$.
Every bit permutation is a bijection on the set $\{0,1\}^n$. Equivalently: every bit permutation is a permutation of the set of $n$-bit bistrings. There are $n!$ such bit permutations, versus the much larger $(2^n)!$ such permutations. The subset of bit permutations is closed under function composition: applying any two fixed bit permutations yields a bit permutation.
Note: some authors use bit transposition or even transposition to designate a bit permutation as defined above, and distinguish it from a permutation. That's not what I'll do in the following, but it could be what the OP has in mind.
A bit transposition is a particular bit permutation of $n$-bit bitstrings, entirely defined by two integers $\ell$ and $c$ with $n=\ell\cdot c$. It's $n$ integers $p_i$ are $p_i=1+((i-1)\bmod c)\cdot c+\lfloor (i-1)/c\rfloor$, for $1\le i\le n$.
In other words, when we write the input bits as $\ell=n/c$ lines and $c$ columns, and the output bits as $c$ lines and $\ell$ columns, output column $j$ comes from input line $j$.
For example, with $\ell=3$ and $c=2$, the bit transposition has $p_1=1$, $p_2=3$, $p_3=5$, $p_4=2$, $p_5=4$, $p_6=6$, which is best represented as
1 3 5
2 4 6
Sometime, the bit permutation with $p_i=\ell\cdot c-((i-1)\bmod c)\cdot c-\lfloor (i-1)/c\rfloor$ is also considered an alternate bit transposition. With $\ell=3$ and $c=2$, the alternate bit transposition has $p_i$
6 4 2
5 3 1
When $n$ is a square and $\ell,c$ are unspecified, they are assumed to be $\sqrt n$, and such square bit transpositions are involutions.
Would the halving of the initial block and eventual swapping of the halves be a permutation or a transposition?
In DES, the transformation (consisting of IP followed by halving) from 64-bit input to LR is a bit permutation (thus a permutation of the set $\{0,1\}^n$, but that's much less specific). It's $p_i$ are given in the IP table:
58 50 42 34 26 18 10 2
60 52 44 36 28 20 12 4
62 54 46 38 30 22 14 6
64 56 48 40 32 24 16 8
57 49 41 33 25 17 9 1
59 51 43 35 27 19 11 3
61 53 45 37 29 21 13 5
63 55 47 39 31 23 15 7
A very regular swapping of lines makes it
64 56 48 40 32 24 16 8
63 55 47 39 31 23 15 7
62 54 46 38 30 22 14 6
61 53 45 37 29 21 13 5
60 52 44 36 28 20 12 4
59 51 43 35 27 19 11 3
58 50 42 34 26 18 10 2
57 49 41 33 25 17 9 1
which is the alternate square bit transposition for $n=64$. Thus IP is nearly a bit transposition.
When seen as operating on LR, the swapping of the halves is a bit permutation, which $p_i$ are given by:
33 34 35 36 37 38 39 40
41 42 43 44 45 46 47 48
49 50 51 52 53 54 55 56
57 58 59 60 61 62 63 64
1 2 3 4 5 6 7 8
9 10 11 12 13 14 15 16
17 18 19 20 21 22 23 24
25 26 27 28 29 30 31 32
It's not a bit transposition, but it's a very regular bit permutation (as bit transpositions are), and an involution (which square bit transpositions are).
