Your Markov chain is a Gaussian $\text{AR}(1)$ process with unit error variance:
$$X_{t+1} = \phi X_t + \varepsilon_{t+1}
\quad \quad \quad \varepsilon_t \sim \text{IID N}(0,1).$$
The elements in this process can be written as weighted sums of normal error terms (in $\text{MA}(\infty)$ form), so the stationary distribution (if it exists) will be a normal distribution, $X_t \sim \text{N}(\mu, \sigma^2)$.
It remains only to find a mean and variance parameter that yield a stationary distribution. To do this, we apply the recursive equation to give equations for the mean and variance parameters. Under the condition of stationarity we have a fixed mean, which means gives the mean equation:
$$\mu = \mathbb{E}(X_{t+1})
= \mathbb{E}(\phi X_t + \varepsilon_{t+1})
= \phi \mathbb{E}(X_t) + \mathbb{E}(\varepsilon_{t+1})
= \phi \mu.$$
Since $-1 < \phi < 1$ this yields the unique solution $\mu = 0$. Under the condition of stationarity we also have a fixed variance, which means gives the variance equation:
$$\sigma^2 = \mathbb{V}(X_{t+1})
= \mathbb{V}(\phi X_t + \varepsilon_{t+1})
= \phi^2 \mathbb{V}(X_t) + \mathbb{V}(\varepsilon_{t+1})
= 1 + \phi^2 \sigma^2.$$
This equation yields the solution $\sigma^2 = 1/(1-\phi^2)$. Thus, the stationary distribution exists, and it is:
$$X_t \sim \text{N} \Bigg( 0, \frac{1}{1-\phi^2} \Bigg).$$
Note that this form is well-known as the stationary distribution of an $\text{AR}(1)$ process with unit variance on the noise terms. It can easily be generalised to broader $\text{AR}(1)$ processes with a non-zero mean, or non-unit variance for the error term.
