Let $(i,j)$ represent the cell in the $i$th row and $j$th column. Let $S_1 = (1,1)$. For $k\geq 2$, let $S_k$ represent the sequence $(1,k),(k,k),(k,k-1),(k-1,k),(k,k-2),(k-2,k), \cdots (k,2),(2,k),(k,1)$. Then $S_1S_2\cdots S_nS_1$ is a possible sequence of $n^2$ moves.

It can be solved pretty easilly by induction. Let for $n$ the statement be true. Then for $n+1$we get: $(k,i)-$ induction $-(j,k)-(k,n+1)-(n+1,1)-(1,n+1)-...-(n+1,k-1)-(k-1,n+1)-(n+1,k+1)-(k+1,n+1)-...-(n+1,n)-(n,n+1)-(n+1,n+1)-(n+1,k)-(k,i)$

Let $(i,j)$ denote the cell at the intersection of the $i^{\text{th}}$ column and $j^{\text{th}}$ row. At first we shall prove that there exists a sequence of $n^2-1$ moves taking the pawn from $(1,1)$ to $(1,n)$ and visiting each cell exactly once. Let us carry out the proof by induction. The base case $n=1$ is trivial. The case $n=2$ is solved by the following example: [asy][asy] import graph; size(3.647326426690823cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-1.6591808684145768,xmax=1.9881455582762464,ymin=2.5285063226884015,ymax=4.698972242840773; pair A=(-0.54,4.46), B=(-0.54,3.7), C=(0.22,3.7), D=(0.22,4.46), F=(-0.22,4.08), G=(0.68,4.08), H=(0.68,3.26), I=(-0.22,3.26); D(A--B--C--D--cycle,white); D(A--B); D(A--B,white); D(B--C,white); D(C--D,white); D(D--A,white); D(A--B); D(A--D); D(C--D); D(D--(0.98,4.46)); D((0.98,4.46)--(0.98,2.94)); D(B--(0.98,3.7)); D(B--(-0.54,2.94)); D((-0.54,2.94)--(0.22,2.94)); D((0.98,2.94)--(0.22,2.94)); D(C--(0.22,2.94)); D(F--G--H--I); label("$(1,2)$",(-0.42767737893682145,3.245703757173533),SE*lsf,fp); label("$(2,1)$",(0.5065666475635446,4.359247142295184),SE*lsf,fp); D(F--G,EndArrow(6)); D(G--H,EndArrow(6)); D(H--I,EndArrow(6)); label("$(1,1)$",(-0.2672516370125162,4.35452873812094),SE*lsf,fp); label("$(2,2)$",(0.47825622251807903,3.245703757173533),SE*lsf,fp); D(A,linewidth(2.pt)); D(B,linewidth(2.pt)); D((-0.54,2.94),linewidth(2.pt)); D(C,linewidth(2.pt)); D(D,linewidth(2.pt)); D((0.98,4.46),linewidth(2.pt)); D((0.98,3.7),linewidth(2.pt)); D((0.98,2.94),linewidth(2.pt)); D((0.22,2.94),linewidth(2.pt)); D(F,linewidth(4.pt)+blue); D(G,linewidth(4.pt)+blue); D(H,linewidth(4.pt)+blue); D(I,linewidth(4.pt)+blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Now suppose we have proved the case $n=k$. For $n=k+1$, firstly cover the top right $k\times k$ square by the induction hypothesis, finishing at $(1,k)$. Now the following valid sequence of moves takes us to $(1,k+1)$: $$\begin{array}{llll} (1,k)\to & (k+1,1)\to & (2,k+1)\to & (k+1,2)\to \\ (3,k+1)\to & (k+1,3)\to & (4,k+1)\to & (k+1,4)\to \\ \;&\cdots & \cdots & \;\\ (k,k+1)\to & (k+1,k)\to & (k+1,k+1)\to & (1,k+1).\end{array}$$This completes the induction step, thereby proving our claim. Returning to the original problem, we see that we can use the above result to reach $(1,n)$ from $(1,1)$ in $n^2-1$ steps, and then return to $(1,1)$, making a total of $n^2$ moves. Since this creates a cycle passing through all cells, we can start at any cell, and follow this cycle to end up at the starting point. This completes our proof. $\blacksquare$