Is a canonical transformation equivalent to a transformation that preserves volume and orientation?

In dimension $2n>2$ they are not equivalent since (for time-independent transformations) canonical is equivalent to $$\sum_{k=1}^n dq^k\wedge dp_k = \sum_{k=1}^n dQ^k\wedge dP_k\tag{1}$$ whereas conservation of oriented volume means $$dq^1\wedge \cdots \wedge dq^n \wedge dp_1\cdots \wedge dp_n = dQ^1\wedge \cdots \wedge dQ^n \wedge dP_1\cdots \wedge dP_n\:.\tag{2}$$ The former is much more restrictive. The latter only requires that the Jacobian matrix has determinant $1$. Already with $4\times 4$ matrices there are easy conterexamples.

$Q^1 = aq^1$,

$Q^2= b q^2$,

$P_1 = b^{-1} p_1$,

$P_2 = a^{-1}p_2$

where coordinates are over $\mathbb R^4$ and with constants $a,b>0$ satisfying $a\neq b$. This transformation satisfies (2) but not (1).

Instead, for $2n=2$, (1) and (2) are evidently equivalent.

1. Counterexample: The transformation $$Q^1~=~2q^1 ,\qquad P_1~=~p_1,\qquad Q^2~=~\frac{1}{2}q^2 ,\qquad P_2~=~p_2 $$ preserves phase space volume & orientation, but is not a symplectomorphism.$^1$

2. For 2D phase space, the canonical phase space volume form $$\Omega~=~\frac{1}{n!}\omega^{\wedge n}$$ is the symplectic 2-form $\omega$ itself, so that the orientation & volume preserving transformations are the symplectomorphisms.

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$^1$ Here we will assume that OP defines a canonical transformation (CT) as a symplectomorphism. Be aware that several non-equivalent definitions of CTs appear in the literature, cf. e.g. this Phys.SE post.