If $f(1)=1$, then is it true that $f(n)=n$ for all $n \in \mathbb{N}\cup\{0\}$.

Yes.

Suppose you already know that $f(n)=n$ for all $n < N$. You want to find $a,b,c$, all smaller than $N$, such that $N^2+a^2=b^2+c^2$. If you find such a triplet, you immediately conclude that $f(N)=N$, and you have the necessary inductive step.

For odd $N>6$, use

$N^2+\left(\frac{N-5}{2}\right)^2 = (N-2)^2+\left(\frac{N+3}{2}\right)^2$

For even $N>6$, use

$N^2+\left(\frac{N}{2}-5\right)^2 = (N-4)^2 + \left(\frac{N}{2}+3\right)^2$

Since the question details already prove that $f(n)=n$ for small values of $n$, this inductive step finishes the argument.

Let $N$ be the smallest number with $f(N)\neq N$.

Notice that $f(N^2+a^2)=f(N)^2 + a^2 \neq N^2 + a^2$ for all $a<N$.

On the other hand if we can write $N^2+a^2$ as $b^2 + c^2$ with $b,c<N$ then $f(b^2+c^2)=f(b)^2+f(c)^2=b^2+c^2$.

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We now prove that for large $N$ we can always find $0\leq a,b,c < N$ such that $N^2+a^2=b^2+c^2$ or $N^2-c^2=(a+b)(a-b)$.

Letting $c=(N-k)$ we get $k(2N-k)=(a+b)(a-b)$.

Checking with $k=1,2,3$ we'll find a value such that $2N-k$ that is a multiple of $3$, this will give rise to a factorization $(\frac{2N-K}{3})(3k)$ which can be realized in the way $(a+b)(a-b)$ with $0\leq a,b<N$.

The first value that makes $N$ large enough is the first value such that $3\times 3 \leq (2N-3)$ which is $6$.

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So we just need to prove $f(n)=n$ holds for $n\in \{0,1,2,\dots,6\}$, we prove them in this order:

$0$ (clearly)

$1$ (clearly)

$2$ (sum of two squares)

$4$ (a square)

$5$ (sum of $1^2+2^2$)

$3$ ($3^2+4^2=5^2$)

$8$ ($2^2+2^2$)

$10$ ($1^2+3^2$)

$6$ ($6^2+8^2=10^2$)