Chebyshev equation for shifted Chebyshev polynomials of the first kind

The Chebyshev polynomials of the first kind ${\displaystyle T_{n}}(x)$ are given by the solutions of the following equation: $${\displaystyle (1-x^{2})y''-xy'+n^{2}y=0,} \quad \quad (1)$$ i.e. $y(x)=T_n(x)$. I would like to know how this equation changes if we consider $T_n\left(\frac{x^2}{2}-1\right)$ instead of $T_n(x)$. Just replace $x$ with $\frac{x^2}{2}-1$ in (1) or not?

It depends...

If you consider a point $u =\frac{x^2}{2}-1$, but still study the map $x \mapsto T_n\left(\frac{x^2}{2}-1\right)$, then yes, you can replace $x$ with $u =\frac{x^2}{2}-1$ in equation $(1)$.

But if your question is related to the map $R_n(x) = T_n\left(\frac{x^2}{2}-1\right)$, then you have to take the derivatives of the compositions

$$\begin{aligned} R_n^\prime(x)=&x T_n^\prime\left(\frac{x^2}{2}-1\right)\\ R_n^{\prime \prime}(x)=& T_n^\prime\left(\frac{x^2}{2}-1\right) + x^2 T_n^{\prime\prime}\left(\frac{x^2}{2}-1\right) \end{aligned}$$ and use those relations in equation $(1)$.