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Suppose we know that $F_X(x)=P(X\le x)=x$. So, if we were to look for $P(X\le x/n)$, can we say that $P(X\le x/n)=\frac{1}{n} P(X\le x)=\frac{1}{n}F_X(x)=\frac{x}{n}$ given $n$ is some constant

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    $\begingroup$ You cannot have $F_X(x)=x$ for all real $x$. Edit the question to make it more precise. $\endgroup$
    – Kavi Rama Murthy
    Commented Jan 18, 2022 at 9:21
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    $\begingroup$ If $F_X(x)=P(X\le x)=x$ when $0 \le x \le 1$ then for $n\ge 1$ it follows immediately that $P(X\le x/n)=\frac{x}{n}$ when $0 \le x \le n$ since $0 \le \frac x n \le 1$ $\endgroup$
    – Henry
    Commented Jan 18, 2022 at 9:40

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No you can't factor out the $\frac1n$ in the CDF, the pdf although of the RV $Y=nX$ has a density of $f_Y(y)=\frac d{dy}F_X(y/n)=\frac1n f_X(y/n)$.

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