I do not know if this is the best way to do this, but I tried to determine $T$ from the given data points $T(p_i) = q_i$.
I got the matrix (see appendix)
$$
T =
\begin{pmatrix}
a-1 & a-2 & a & a \\
b-11 & b-10 & b-6 & b \\
c-1 & c & c-1 & c \\
d & d & d & d
\end{pmatrix}
$$
where $a,b,c,d$ are free variables. So we have many candidates for $T$.
Solving for $T x = 0$:
$$
\begin{pmatrix}
a-1 & a-2 & a & a \\
b-11 & b-10 & b-6 & b \\
c-1 & c & c-1 & c \\
d & d & d & d
\end{pmatrix}
\to
\begin{pmatrix}
a-1 & a-2 & a & a \\
b-11 & b-10 & b-6 & b \\
c-1 & c & c-1 & c \\
1 & 1 & 1 & 1
\end{pmatrix}
\to
\\
\begin{pmatrix}
a & a-1 & a+1 & a+1 \\
b & b+1 & b+5 & b+11 \\
c & c+1 & c & c+1 \\
1 & 1 & 1 & 1
\end{pmatrix}
\to
\begin{pmatrix}
0 & -1 & 1 & 1 \\
0 & 1 & 5 & 11 \\
0 & 1 & 0 & 1 \\
1 & 1 & 1 & 1
\end{pmatrix}
\to
\\
\begin{pmatrix}
0 & 0 & 1 & 2 \\
0 & 0 & 5 & 10 \\
0 & 1 & 0 & 1 \\
1 & 0 & 1 & 0
\end{pmatrix}
\to
\begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 2 \\
0 & 1 & 0 & 1 \\
1 & 0 & 0 & -2
\end{pmatrix}
\to
\begin{pmatrix}
1 & 0 & 0 & -2 \\
0 & 1 & 0 & 1 \\
0 & 0 & 1 & 2 \\
0 & 0 & 0 & 0
\end{pmatrix}
$$
This gives the solutions $x = (2s, -s, -2s, s)^\top = s (2,-1,-2,1)^\top = s (2 -t -2 t^2 + t^3)$ for $s \in \mathbb{R}$, thus $\ker T$ is created by one basis vector, which agrees with your solution.
Determining $T$
We know $Q = T P$ with
P =
-1 0 -1 0
1 -1 0 2
0 1 0 -1
0 0 1 -1
Q =
-1 2 1 -4
1 4 11 -14
1 -1 1 1
0 0 0 0
and then linearize $T$ into $x = (t_{11}, t_{12}, t_{13}, t_{14}, t_{21}, \dotsc, t_{44})^\top$ to get a system $A x = b$ with
A =
-1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
-1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
0 2 -1 -1 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 -1 1 0 0 0 0 0 0 0 0 0
0 0 0 0 -1 0 0 1 0 0 0 0 0 0 0 0
0 0 0 0 0 2 -1 -1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 -1 1 0 0 0 0 0
0 0 0 0 0 0 0 0 -1 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 0 2 -1 -1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1 0
0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 2 -1 -1
and
b =
-1
2
1
-4
1
4
11
-14
1
-1
1
1
0
0
0
0
Then Gauss-Jordan elimination gives the row-echelon form:
>> rref([A,b])
ans =
Columns 1 through 15:
1 0 0 -1 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0
0 1 0 -1 0 0 0 0 0 0 0 0 0 0 0
0 0 1 -1 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0 -0
0 0 0 0 1 0 0 -1 -0 -0 -0 -0 -0 -0 -0
0 0 0 0 0 1 0 -1 0 0 0 0 0 0 0
0 0 0 0 0 0 1 -1 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0 0 -1 -0 -0 -0
0 0 0 0 0 0 0 0 0 1 0 -1 0 0 0
0 0 0 0 0 0 0 0 0 0 1 -1 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
and
Columns 16 and 17:
-0 -1
0 -2
-0 -0
-0 -11
0 -10
0 -6
-0 -1
0 0
0 -1
-1 -0
-1 0
-1 0
0 0
0 0
0 0
0 0