Least permutations needed to permute from decreasing order to increasing order You are asked to permute the neighboring sub-sequence of the sequence $n,n-1,n-2,\cdots,1$ until the sequence is brought to the increasing order.  
By permute the neighboring sub-sequence I mean for example:
$5,4,3,2,1 \to 5,3,4,2,1 $ or $5,4,3,2,1\to 5,2,4,3,1$ or $5,4,3,2,1\to5,2,1,4,3$. 
What is the least number of permutations needed?
Edit
A first nontrivial case I can come up with is:
$$
54321\to52143\to14523\to12345
$$
Edit 2
A second nontrivial case I come up with to show $T(n)<f(n)$ is possible regarding the answer by @Brian M. Scott:
$$
7654321\to7632154\to7215634\to1567234\to1234567
$$
maybe $\frac{n+1}{2}$?

You need at least $\lceil (n+1)/2\rceil.$ Consider the number $G$ of terms whose right neighbor is greater. $G$ starts at zero and ends as $n-1.$
For example
$$G(5,4,3,2,1)=0$$
$$G(5,2,1\color{red},4,3)=1$$
$$G(1\color{red},4\color{red},5,2\color{red},3)=3$$
$$G(1\color{red},2\color{red},3\color{red},4\color{red},5)=4$$
I have colored in red the comma between each term and its right neighbour whenever that neighbour is greater.
By Brian M. Scott's definition given in the comments, each swap is of the form $WXYZ\to WYXZ$ where $W,X,Y,Z$ are possibly empty subwords. Let $q,r,s,t$ be the lengths of $W,X,Y,Z$ respectively. Let
$$w_1,\dots,w_q,x_1,\dots,x_r,y_1,\dots,y_s,z_1,\dots,z_t=WXYZ$$
so
$$w_1,\dots,w_q,y_1,\dots,y_s,x_1,\dots,x_r,z_1,\dots,z_t=WYXZ.$$
For example for $5,4,3,2,1\to 5,2,1,4,3$ we have $(q,r,s,t)=(1,2,2,0).$
Note
\begin{align*}
G(WYXZ)-G(WXYZ)=[w_q<y_1]+[y_s<x_1]+[x_r<z_1]\\-[w_q<x_1]-[x_r<y_1]-[y_s<z_1]\tag{1}
\end{align*}
where $[a<b]=1$ if $a<b$ and $[a<b]=0$ otherwise (Iverson bracket notation). If $q=0$ ignore the terms using $w_q,$ and if $t=0$ ignore the terms using $z_1.$
We can write the entire sequence of swaps as $P_0\to P_1\to \dots\to P_m$ where $m$ is the number of swaps, and $P_0=(n,n-1,\dots,1),$ and $P_m=(1,2,\dots,n).$
In the first swap $WXYZ=P_0\to P_1=WYXZ,$ we always have $w_q>y_1$ and $x_r>z_1.$ So by (1) we must have $G(P_1)-G(P_0)\leq 1.$ 
In the final swap $WXYZ=P_{m-1}\to P_m=WYXZ$ we always have $w_q<x_1$ and $y_s<z_1,$ so $G(P_m)-G(P_{m-1})\leq 1.$
Now consider an arbitrary swap $WXYZ=P_{i}\to P_{i+1}=WYXZ.$ From (1) we know $G(P_{i+1})-G(P_i)\leq 3,$ and it cannot be exactly $3$ because that would give $w_q<y_1<x_r<z_1<y_s<x_1<w_q.$ So $G(P_{i+1})-G(P_i)\leq 2.$ To summarize:
\begin{align*}
&G(P_{1})&-&&&G(P_0)&\leq 1\\
&G(P_{i+1})&-&&&G(P_i)&\leq 2&\qquad\text{ for }1\leq i\leq m-2\\
&G(P_{m})&-&&&G(P_{m-1})&\leq 1
\end{align*}
which sum to $n-1=G(P_m)-G(P_0)\leq 2m-2.$ So $m\geq\lceil (n+1)/2\rceil.$
