One, (40 points) Find all real-coefficient polynomials $p(x)$ satisfying
$$
\begin{array}{l}
\left(x^{3}+2 x^{2}+3 x+2\right) p(x-1) \\
=\left(x^{3}-4 x^{2}+5 x-6\right) p(x)
\end{array}
$$
Given conditions are equivalent to
$$
\begin{array}{l}
(x+1)\left(x^{2}+x+2\right) p(x-1) \\
=(x-3)\left(x^{2}-x+2\right) p(x) .
\end{array}
$$

Substituting $x=-1, x=3$ into equation (1) yields
$$
p(-1)=0, p(2)=0 \text {; }
$$

Substituting $x=0, x=2$ into equation (1) yields
$$
p(0)=0, p(1)=0 \text {. }
$$

Therefore, we can assume $p(x)=x(x-1)(x+1)(x-2) q(x)$.
Substituting into equation (1) gives
$$
\begin{aligned}
& (x+1)\left(x^{2}+x+2\right)(x-1)(x-2) x . \\
& (x-3) q(x-1) \\
= & (x-3)\left(x^{2}-x+2\right) x(x-1)(x+1) . \\
& (x-2) q(x) \\
\Rightarrow & \left(x^{2}+x+2\right) q(x-1) \\
& =\left(x^{2}-x+2\right) q(x) .
\end{aligned}
$$

Since $\left(x^{2}+x+2, x^{2}-x+2\right)=1$, we have
$$
q(x)=\left(x^{2}+x+2\right) r(x) \text {. }
$$

Substituting into equation (2) gives
$$
\begin{array}{l}
\left(x^{2}+x+2\right)\left(x^{2}-x+2\right) r(x-1) \\
=\left(x^{2}-x+2\right)\left(x^{2}+x+2\right) r(x) .
\end{array}
$$

The above equation always holds, indicating that $r(x)$ is a constant $C$. Therefore,
$$
q(x)=C\left(x^{2}+x+2\right) \text {. }
$$

Thus, the desired polynomial is
$$
p(x)=C\left(x^{2}+x+2\right) x(x-1)(x+1)(x-2) \text {, }
$$

where $C$ is a constant.