Can $Y_t= X_t - t X_1 -(1-t)X_0$ be Gaussian without $(X_t)$ being Gaussian?

Let $$Y_t= X_t - t X_1 -(1-t)X_0$$ be a sample-continuous stochastic process defined on $$[0,1]$$. If $$(Y_t)$$ is Gaussian, does it imply that $$(X_t)$$ is Gaussian on $$(0,1)$$ ?

It feels wrong but I am unable to construct a counterexample.

Here are some preliminaries:

A time continuous stochastic process $$\left\{X_{t} ; t \in T\right\}$$ is Gaussian if and only if for every finite set of indices $$t_{1}, \ldots, t_{k}$$ in the index set $$T$$, $$\mathbf{X}_{t_{1}, \ldots, t_{k}}=\left(X_{t_{1}}, \ldots, X_{t_{k}}\right)$$

is a multivariate Gaussian random variable. That is the same as saying every linear combination of $$\left(X_{t_{1}}, \ldots, X_{t_{k}}\right)$$ has a univariate Gaussian distribution.

Using characteristic functions of random variables, the Gaussian property can be formulated as follows: $$\left\{X_{t} ; t \in T\right\}$$ is Gaussian if and only if, for every finite set of indices $$t_{1}, \ldots, t_{k}$$, there are real-valued $$\sigma_{\ell j}, \mu_{\ell}$$ with $$\sigma_{j j}>0$$ such that the following equality holds for all $$s_{1}, s_{2}, \ldots, s_{k} \in \mathbb{R}$$ $$\begin{equation*} \mathrm{E}\left(\exp \left(i \sum_{\ell=1}^{k} s_{\ell} \mathbf{X}_{t_{\ell}}\right)\right)=\exp \left(-\frac{1}{2} \sum_{\ell, j} \sigma_{\ell j} s_{\ell} s_{j}+i \sum_{\ell} \mu_{\ell} s_{\ell}\right) . \end{equation*}$$ where $$i$$ denotes the imaginary unit such that $$i^{2}=-1$$. The numbers $$\sigma_{\ell j}$$ and $$\mu_{\ell}$$ can be shown to be the covariances and means of the variables in the process.

Here is a counterexample: Suppose, that $$X_t = W_t + tZ,$$ where $$W_t$$ is a Gaussian proces and $$Z$$ is any non-Gaussian random variable (independent of $$W_t$$), then $$X_t$$ is not Gaussian, however \begin{align*} Y_t &= X_t - tX_1 - (1-t)X_0 \\ &= (W_t + tZ) - t(W_1+Z) - (1-t)(W_0+0) \\ &= W_t - tW_1 - (1-t)W_0 \end{align*} is indeed a Gaussian proces.