This is possible for all $n$ and $p$.

I start with a **direct construction**.

Obviously, if $X$ is even, then we should have $X'=X$. So we should construct the corresponding coupling between $Y$ and $X'$, where $Y$ is the $B(n,p)$ restricted to odd outcomes.

Choose $2n$ i.i.d. Bernoulli$(p)$ variables $\xi_1,\ldots,\xi_n;\eta_1,\ldots,\eta_n$ and condition to $$\sum (\xi_i+\eta_i) \quad \text{is odd}.$$

Denote by $\Omega$ the set of possible $2^{2n-1}$ outcomes and consider the map $\Phi:\Omega\to \Omega$: choose the minimal $i$ for which $\xi_i\ne \eta_i$ and switch $\xi_i$ and $\eta_i$. This is a measure-preserving involution. Note that $\Phi$ changes the parity of $S=\eta_1+\ldots+\eta_n$, so $S$ is even with probability $1/2$. Next, if we further condition to ($S$ is even), then $S$ becomes distributed as $X'$. Indeed, this clearly holds even we fix all $\xi_i$'s (with odd sum). Analogously, if $S$ is odd, it is distributed as $Y$.

Now our coupling: choose $\omega\in \Omega$ at random, set $\{X',Y\}=\{S(\omega), S(\Phi(\omega)\}$.

Well, now goes a boring explanation **how to get this coupling** using generating functions.

Let $c_0,c_2,\ldots$ be probabilities of outcomes $0,2,\ldots$ for $X'$, we have $c_0+c_2x^2+\ldots=\frac{(q+px)^n+(q-px)^n}{1+\delta^n}$, $\delta:=q-p$ (and $q=1-p$). Denote the probabilities for $Y$ by $c_1,c_3,\ldots$, then $c_1+c_3x^2+\ldots=\frac{(q+px)^n-(q-px)^n}{1-\delta^n}$.

How may our coupling between $Y$ and $X$ look like? There is no freedom: if $Y=1$, then $X'\in \{0,2\}$ with probabilities corr. $c_0$ and $c_1-c_0$ (these are not conditional probabilities, I mean, $c_1-c_0={\rm prob}(Y=1,X'=2)$ etc.) If $Y=3$, then $X'\in \{2,4\}$ with probabilities $c_2-c_1+c_0$ and $c_3-c_2+c_1-c_0$, etc. Thus what we need is that all alternating sums $c_k-c_{k-1}+c_{k-2}-\ldots$ must be non-negative, or: all coefficients of
$$
F(x):=(c_0+c_1x+c_2x^2+\ldots)(1-x+x^2+\ldots)
$$
must be non-negative.
We have
$$
F(x)=2\frac{(q+px)^n-\delta^n(q-px)^n}{(1+x)(1-\delta^{2n})}=
2\frac{((q+p)(q+px))^n-((q-p)(q-px))^n}{(1+x)(1-\delta^{2n})}=\\
2\frac{((q^2+p^2x)+pq(1+x))^n-((q^2+p^2x)-pq(1+x))^n}{(1+x)(1-\delta^{2n})},
$$
and expanding $((q^2+p^2x)\pm pq(1+x))^n$ by Binomial we see that $F(x)$ is indeed a polynomial with non-negative coefficients.