If a newform $f \in \mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N),\varepsilon)$ has an inner twist by some $\sigma \in \operatorname{Aut}(\mathbb{C})$, then $f^{\sigma}$ is a newform of the same level as $f$. Moreover, if $\varepsilon$ is trivial, then so is the nebentypus of $f^{\sigma}$ (see (3.8) of Ribet's paper), and so any inner twist must arise from a quadratic Dirichlet character.

Ribet's paper actually shows (see (3.9)) that if $N$ is squarefree, then there are no nontrivial inner twists. More precisely, he shows that if $N$ is squarefree and $\chi$ is a quadratic Dirichlet character, then the twist $f \otimes \chi$ cannot have level $N$ and trivial nebentypus.

This is not hard to see by a local argument: if $N$ is squarefree, then for any $p \mid N$, the local component $\pi_p$ of the associated automorphic representation must be isomorphic to $\mathrm{St}$, the Steinberg representation of conductor $p$ associated to the trivial character. Any nontrivial twist of $f$ leaving the nebentypus unchanged must necessarily be quadratic, but any twist of $\mathrm{St}$ by a ramified quadratic character has conductor at least $p^2$, not $p$ (and in fact exactly $p^2$ if $p$ is odd). Similarly, at any unramified place, the twist of an unramified principal series representation by a ramified quadratic character has conductor at least $p^2$. So the level of $f \otimes \chi$ is strictly greater than $N$.

But if $N$ is not squarefree, then this is no longer the case! Indeed, in a recent paper (Theorem 6.4), I proved the following result (well, technically I proved it for Maaß cusp forms, but it generalises in an obvious way to holomorphic cusp forms).

Fix a nonsquarefree odd integer $N$, and let $N' > 1$ be squarefree and such that $N'^2 \mid N$. Let $\mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))_{\mathrm{nonmon}(\varepsilon_{\mathrm{quad}(N')})}$ denote the vector space spanned by newforms $f$ of weight $k$, level $N$, and trivial nebentypus such that $f$ does not have CM by the quadratic Dirichlet character $\varepsilon_{\mathrm{quad}(N')}$ modulo $N'$, but that the twist $f \otimes \varepsilon_{\mathrm{quad}(N')}$ is also a newform of level $N$ and trivial nebentypus. Then
\[\frac{\dim \mathcal{S}_{k}^{\mathrm{new}} (\Gamma_0(N))_{\mathrm{nonmon} (\varepsilon _{\mathrm{quad}(N')} )} }{\dim \mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))} \sim \prod_{\substack{p \mid N' \\ p^2 \parallel N}} \left(1 - \frac{p}{p^2 - p - 1}\right)\]
as $k$ tends to infinity over the even integers.

Furthermore, this still holds if we replace $\mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))_{\mathrm{nonmon}(\varepsilon_{\mathrm{quad}(N')})}$ with
\[\bigcap_{\substack{N^* \mid N' \\ N^* > 1}} \mathcal{S}_k^{\mathrm{new}}(\Gamma_0(N))_{\mathrm{nonmon}(\varepsilon _{\mathrm{quad}(N^*)})}.\]

A similar result holds even when $N$ is squarefree if $\varepsilon$ is nontrivial; see Proposition 6.5 of my paper.

That inner twists occur in abundance when $N$ is not squarefree was observed as far back as 1977 by Ribet; at the bottom of page 48 of this paper, Ribet writes

It would be of interest to give an *a priori* construction of forms
with extra twists. If the level $N$ is divisible by a high power of a
prime, these forms seem to be more the rule than the exception.