No, the two coherent states are not equivalent, exactly due to the reason mentioned by Arnold. They do not carry the same representation of $SO(2N)$.

The first coherent state $|\mathbf{\gamma} \rangle$ is a coherent state of the Berezin-Schwinger group $\mathfrak{bs} = \mathbb{R}^{0, 2n} \rtimes \mathbb{R}^{1, 0}$, which is the fermionic counterpart of the Weyl-Heisenberg group $\mathfrak{hw}$. It is built as the orbit of $ \mathfrak{bs} $ through the fermionic Fock vacuum isomorphic to the coset space $\mathbb{R}^{0, 2n} = \mathfrak{bs} / \mathbb{R}^{1, 0}$. This coherent state lives in a Hilbert superspace spanned by a $2^N$ dimensional basis.

The second coherent state is $|\mathbf{\xi} \rangle$ a coherent state of $Spin(2N)$, it is a coherent state representation of one of the spinor representations of this group. It is built as the orbit of $SO(2N)$ through the Fock vacuum which is isomorphic to the coset space $SO^*(2N)/P = SO(2N)/U(N)$ .

It is true that the dimensions of the representation spaces are the same. But the coherent states are not the same. The reason is that a coherent state irreducibly represents the group which acts on the vacuum. In the first it is the $ \mathfrak{bs} $ group and in the second it is $SO(2N)$.

It is not hard to see in the second case that the Cartan generators of SO(2N) are represented by:

$$H_i = a_i^{\dagger} a_i - \frac{1}{2}$$

Thus the weights of the vacuum are:

$$\begin{bmatrix} -\frac{1}{2},&-\frac{1}{2} &...&-\frac{1}{2} \end{bmatrix}^t$$

Which are exactly the weights of the lowest weight of the spinor representation.

The group $SO(2N)$ acts on the coherent state manifold of the $ \mathfrak{bs} $ group as

$$\begin{bmatrix} a& b\\ \bar{b}& \bar{a} \end{bmatrix} \begin{bmatrix} \mathbf{c}\\ \mathbf{\bar{c}} \end {bmatrix}$$

(The matrix representation is in a $U(2N)$ basis), and:

$$\mathbf{c} = \begin{bmatrix} c_1,&... &c_N \end{bmatrix}^t$$

Thus the action of $SO(2N)$ on the coherent state manifold is homogeneous, therefore the vacuum is invariant. Moreover, it is not hard to see that the action of $SO(2N)$ on the coherent states spanned by a partial subset of Grassmann coordinates, for example those in which only $\gamma_1$ and $\gamma_2$ are nonvanishing is closed, because it cannot change the ghost number of the terms.

Thus the action decomposes to a direct sum

$$\bigoplus_{k=0}^N \bigwedge_k {N}$$

$N$ is the fundamental representation.

However, in spite what was said above, the two coherent state representations are not unrelated. There is a very interesting series of works by Berceanu in which the bosonic counterpart of:

$$|\mathbf{\gamma} \rangle \otimes |\mathbf{\xi} \rangle$$

was considered, please see for example the following

http://projecteuclid.org/download/pdf_1/euclid.jgsp/1495245699

article and the following works by him citing this work.

In the bosonic case, the combined coherent state represents the group

$$\mathfrak{hw} \rtimes Sp(N)$$

whose fermionic analogy is:

$$\mathfrak{bs} \rtimes SO(2N)$$

In the bosonic case, the $\mathfrak{hw} \rtimes Sp(N)$ coherent state describes a multimode squeezed coherent state.

All the computations in the article can be repeated for the fermionic case with the obvious changes of signs when necessary.

The fermionic coherent state should describe the ground state of some fermionic superfluid.