Consider the mapping $g:t\to 1- at$. It has a fixed point $t_0=1/(1+a)\;$. Denote $g_n$ the $n$-th iteration of $g$. Then $g_n(t)=(-a)^n t+1-a+\ldots+(-a)^{n-1}\;$, $g'_n(t)=(-a)^n$, $n=0,1,\ldots\;$. From the equation we have $$ f^n(t)=x^n g_1'(t)g_2'(t)\ldots g_{n-1}'(t)f(g_n(t)), $$ so $f^n(t_0)=(-a)^{n(n-1)/2}x^n f(t_0)$. The expansion of $f$ in the Taylor series at $t=t_0$ is $$ f(t)=f(t_0)\sum_{n=0}^\infty\frac{(-a)^{n(n-1)/2}x^n(t-t_0)^n}{n!}. $$ The value of $f(t_0)$ can be obtained form the initial condition. For $a=1$ the solution has an explicit form: $$ f(t)=\frac{\cos \left(\left(t-\frac{1}{2}\right) x\right)+\sin \left(\left(t-\frac{1}{2}\right) x\right)}{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}. $$

Consider the mapping $g:t\to 1- at$. It has a fixed point $t_0=1/(1+a)\;$. Denote $g_n$ the $n$-th iteration of $g$. Then $g_n(t)=(-a)^n t+1-a+\ldots+(-a)^{n-1}\;$, $g'_n(t)=(-a)^n$, $n=0,1,\ldots\;$. From the equation we have $$ f^{(n)}(t)=x^n g_1'(t)g_2'(t)\ldots g_{n-1}'(t)f(g_n(t)), $$ so $f^{(n)}(t_0)=(-a)^{n(n-1)/2}x^n f(t_0)$. The expansion of $f$ in the Taylor series at $t=t_0$ is $$ f(t)=f(t_0)\sum_{n=0}^\infty\frac{(-a)^{n(n-1)/2}x^n(t-t_0)^n}{n!}. $$ The value of $f(t_0)$ can be obtained form the initial condition. For $a=1$ the solution has an explicit form: $$ f(t)=\frac{\cos \left(\left(t-\frac{1}{2}\right) x\right)+\sin \left(\left(t-\frac{1}{2}\right) x\right)}{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}. $$

Consider the mapping $g:t\to 1- at$. It has a fixed point $t_0=1/(1+a)\;$. Denote $g_n$ the $n$-th iteration of $g$. Then $g_n(t)=(-a)^n t+1-a+\ldots+(-a)^{n-1}\;$, $g'_n(t)=(-a)^n$, $n=0,1,\ldots\;$. From the equation we have $$ f^n(t)=x^n g_1'(t)g_2'(t)\ldots g_{n-1}'(t)f(g_n(t)), $$ so $f^n(t_0)=(-a)^{n(n-1)/2}x^n f(t_0)$. The expansion of $f$ in the Taylor series at $t=t_0$ is $$ f(t)=f(t_0)\sum_{n=0}^\infty\frac{(-a)^{n(n-1)/2}x^n(t-t_0)^n}{n!}. $$ The value of $f(t_0)$ can be obtained form the initial condition. For $a=1$ the solution has a closed form: $$ f(t)=\frac{\cos \left(\left(t-\frac{1}{2}\right) x\right)+\sin \left(\left(t-\frac{1}{2}\right) x\right)}{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}. $$

Consider the mapping $g:t\to 1- at$. It has a fixed point $t_0=1/(1+a)\;$. Denote $g_n$ the $n$-th iteration of $g$. Then $g_n(t)=(-a)^n t+1-a+\ldots+(-a)^{n-1}\;$, $g'_n(t)=(-a)^n$, $n=0,1,\ldots\;$. From the equation we have $$ f^n(t)=x^n g_1'(t)g_2'(t)\ldots g_{n-1}'(t)f(g_n(t)), $$ so $f^n(t_0)=(-a)^{n(n-1)/2}x^n f(t_0)$. The expansion of $f$ in the Taylor series at $t=t_0$ is $$ f(t)=f(t_0)\sum_{n=0}^\infty\frac{(-a)^{n(n-1)/2}x^n(t-t_0)^n}{n!}. $$ The value of $f(t_0)$ can be obtained form the initial condition. For $a=1$ the solution has an explicit form: $$ f(t)=\frac{\cos \left(\left(t-\frac{1}{2}\right) x\right)+\sin \left(\left(t-\frac{1}{2}\right) x\right)}{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}. $$