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i could not Edit my answer so i have posted new answer

1)let $k<1$ and $f(x)=a^{x}$ in which $a>1$ so

$f'(x)=a^{x}\log a$ and $f(x+k)=a^{x+k}$ then $k=\log \log a/\log a$

2)let $k\ge 1$ and $f(x)=g(x)a^{-x}$ in which, $1<a<1.76$, and $g(x)=-1$, if, $x>1$, $g(x)=1$

if $0<x<1$ ,otherwise $g(x)=0$, so for $0<x<1$,

$f'(x)=-a^{-x}\log a$ and $f(x+k)=-a^{-x-k}$ then $k=-\log \log a/\log a$

i could not Edit my answer so i have posted new answer

1)let $f(x)=a^{x}$ in which $a>1$ so

$f'(x)=a^{x}\log a$ and $f(x+k)=a^{x+k}$ then $k=\log \log a/\log a$ this is hold for $k<1$

2)let $f(x)=g(x)a^{-x}$ in which, $1<a<1.76$, and $g(x)=-1$, if, $x>1$, $g(x)=1$

if $0<x<1$ ,otherwise $g(x)=0$, so for $0<x<1$,

$f'(x)=-a^{-x}\log a$ and $f(x+k)=-a^{-x-k}$ then $k=-\log \log a/\log a$ this is hold for $k\ge 1$

i could not Edit my answer so i have posted new answer

1)let $k<1$ and $f(x)=a^{x}$ in which $a>1$ so

$f'(x)=a^{x}loga$ and $f(x+k)=a^{x+k}$ then $k=logloga/loga$

2)let $k\ge 1$ and $f(x)=g(x)a^{-x}$ in which, $1<a<1.76$  ,and $g(x)=-1$  ,if, $x>1$  ,$g(x)=1$

if $0<x<1$,otherwise $g(x)=0$, so for $0<x<1$

$f'(x)=-a^{-x}loga$ and $f(x+k)=-a^{-x-k}$ then $k=-logloga/loga$

i could not Edit my answer so i have posted new answer

1)let $k<1$ and $f(x)=a^{x}$ in which $a>1$ so

$f'(x)=a^{x}\log a$ and $f(x+k)=a^{x+k}$ then $k=\log \log a/\log a$

2)let $k\ge 1$ and $f(x)=g(x)a^{-x}$ in which, $1<a<1.76$, and $g(x)=-1$, if, $x>1$,  $g(x)=1$

if $0<x<1$  ,otherwise $g(x)=0$, so for $0<x<1$,

$f'(x)=-a^{-x}\log a$ and $f(x+k)=-a^{-x-k}$ then $k=-\log \log a/\log a$

i could not Edit my answer so i have posted new answer

1)let $k<1$ and $f(x)=a^{x}$ in which $a>1$ so

$f'(x)=a^{x}loga$ and $f(x+k)=a^{x+k}$ then $k=logloga/loga$

2)let $k\ge 1$ and $f(x)=g(x)a^{-x}$ in which $1<a<1.76$ and $g(x)=-1 if x>1$ ,$g(x)=1 if 0<x<1$,otherwise $g(x)=0$, so for $0<x<1$

$f'(x)=-a^{-x}loga$ and $f(x+k)=-a^{-x-k}$ then $k=-logloga/loga$

i could not Edit my answer so i have posted new answer

1)let $k<1$ and $f(x)=a^{x}$ in which $a>1$ so

$f'(x)=a^{x}loga$ and $f(x+k)=a^{x+k}$ then $k=logloga/loga$

2)let $k\ge 1$ and $f(x)=g(x)a^{-x}$ in which, $1<a<1.76$ ,and $g(x)=-1$ ,if, $x>1$ ,$g(x)=1$

if $0<x<1$,otherwise $g(x)=0$, so for $0<x<1$

$f'(x)=-a^{-x}loga$ and $f(x+k)=-a^{-x-k}$ then $k=-logloga/loga$