There are functions with f (n) = f (2n) which can't be calculated I have to proofe that there are functions defined by $f:\mathbb{N} \rightarrow \mathbb{N}, f(n)=f(2n), \forall n\in \mathbb{N}$, which are not-computable. However I'm not really sure about the correct method.

I thought about a proof by contradiction. Assume each of those functions are computable. Then, by the Church-turing-thesis, there has to exist a TM which can compute every of those functions. Therefore $L(M)=\{code(M) | \text{M calculates this type of function}\}$ would be decidable. However I profed earlier, that this language is undecidable. This would lead to a contradiction, but I'm not sure about the correctness of my profe...

Thank you for your help :)
Cook up some encoding of Turing machines so that if $n$ codes a machine, $2 n$ codes "the same" (perhaps use binary numbers ending in 1 as starting points, and 0s at the end are disregarded, thus making that odd $n$ and $n \cdot 2^k$ represent the same machine). Then use that e.g. $\operatorname{HALT}(M)$ (does $M$ halt if started on an empty tape?) isn't computable.