The thread at Normal Distribution Existence Non-affine Invariant Transformation? exhibits many non-affine transformations that map normally distributed variables into normally distributed variables, so the answer is in the negative.
(For the record, an affine transformation $f:\mathbb R\to \mathbb R$ is one of the form $f(x) = \mu + \sigma x$ for numbers $\mu$ and $\sigma.$ That is, its graph is a line. I will restrict this definition to positive $\sigma$ (positively sloping graphs) in order to obtain an easily stated result.)
We can, however, point to a simple condition that assures a positive answer, as suggested in the question. I will state it rather generally because that reveals its nature and it is no harder to prove the generalization.
Let $X$ be a continuous random variable supported on all the real numbers. (This implies its distribution function $F_X$ given by $F_X(x)=\Pr(X\le x)$ is a strictly increasing continuous function from $\mathbb R$ onto the interval $(0,1).$)
Denote by $\mathcal{F}[F_X]$ the set of distributions of all invertible affine transformations of $X$ (the "location-scale family" of $X$). Thus
$$\mathcal{F}[F_X] = \{F_{\mu+\sigma X}\mid \mu\in\mathbb{R},\,\sigma\gt 0\}.$$
Proposition
Let $f:\mathbb R\to\mathbb R$ be a continuous, increasing, one-to-one, measurable function for which $F_{f(X)}\in\mathcal{F}[F_X].$ Then $f$ is affine.
Proof
The measurability of $f$ is required to assure $Y=f(X)$ is a random variable.
By applying a preliminary affine transformation to $f(X)$ we may, without any loss of generality, assume $F_X=F_Y=F_{f(X)}.$
Let $y$ be any real number and compute
$$F_X(y) = F_Y(y) = F_{f(X)}(y) = \color{red}{\Pr(f(X) \le y) = \Pr\left(X\le f^{-1}(y)\right)} = F_X\left(f^{-1}(y)\right).$$
The first three and last equalities are all definitions. The remaining (penultimate) equality, shown in red, is justified because $f$ is one-to-one and increasing, whence its inverse $f^{-1}$ exists, is unique, and preserves inequalities.
Finally, comparing the left and right sides of the foregoing, observe that because $F_X$ is strictly increasing and continuous, equality can hold only when the arguments are equal: that is,
$$F_X(y) = F_X\left(f^{-1}(y)\right)\text{ implies } y = f^{-1}(y)$$
for all real $y.$ That means $f^{-1}$ (and therefore $f$) is the identity function. Accounting for the preliminary affine transformation that was applied, we conclude that the original transformation was affine, QED.
Readers might enjoy constructing counterexamples to the proposition when any of the conditions on $f$ are relaxed.
To answer the question: This applies to Normally distributed variables because they are a location-scale family of absolutely continuous variables with distributions supported on the entire real line.
