Partial answer:
Lemma: Let $n=am+1$ where $a\ge1$ and $m\ge2$ are integers. Suppose that $m\mid\phi(n)$ and $a<p$ where $p=\min\{p^*\in\Bbb P:p^*\mid m\}$. If $n$ is not prime then either
$n$ is of the form $\prod p_i$ where $p_i$ are primes, or
$n$ is of the form $2^kr$ where $k,r$ are positive integers.
Proof: Suppose that $n$ is composite. First, note that $m$ must be odd as otherwise, $a=1$ which yields $n-1=m$. The condition $m\mid\phi(n)$ forces $n$ to be prime which is a contradiction.
Next, write $n=q^kr$ where $k,r$ are positive integers and $q$ is a prime such that $(q,r)=1$. As $\phi(n)=q^{k-1}(q-1)\phi(r)$ the condition $m\mid\phi(n)$ yields $$q^{k-1}(q-1)\phi(r)=mt\implies aq^{k-1}(q-1)\phi(r)=t(q^kr-1)$$ for some positive integer $t$. It follows that either $k=1$ or $t=q^{k-1}v$ for some integer $v\ne t$. In the latter case, we obtain $$\frac{q^kr-1}{q^{k-1}(q-1)\phi(r)}=\frac{aps}{mt}=\frac at\implies p>\frac{t(q^kr-1)}{q^{k-1}(q-1)\phi(r)}.$$ Combining this with the trivial result $p<q^{k-1}(q-1)\phi(r)/t$ yields $$t<\frac{q^{k-1}(q-1)\phi(r)}{\sqrt{q^kr-1}}\implies v<\frac{(q-1)\phi(r)}{\sqrt{q^kr-1}}.$$ Substituting back into $n=am+1$ gives $$q^kr-1=\frac av(q-1)\phi(r)\implies aq\phi(r)-vq^kr=a\phi(r)-v>\phi(r)\left(a-\frac{q-1}{\sqrt{q^kr-1}}\right)$$ which is positive since $k\ge2$. This yields $a>vq^{k-1}\ge vq$. Since $p$ is the least prime divisor of $m$, we have $p\le q-1$, unless $q=2$ or $q-1=v$.
Evidently, the first case contradicts $a<p$, so $k=1$. This means that $n$ must be of the form $\prod p_i$ where $p_i$ are primes. The condition $m\mid\phi(n)$ gives $\prod(p_i-1)=bm$ for some positive integer $b$, and substituting this into $n=am+1$ yields $$a=b\frac{\prod p_i-1}{\prod(p_i-1)}.$$ When $m$ is even, we have $a<p\implies a<2$ which implies that $m=\prod p_i-1$. Further, $$b<\frac{2\prod(p_i-1)}{\prod p_i-1}<2\implies m=\prod(p_i-1).$$ The only way that $\prod p_i-1=\prod(p_i-1)$ is when $\prod p_i$ is prime, which solves the problem. Finally, notice that $m$ is odd only when $b=2^{\nu_2(\prod(p_i-1))}d$ for some positive integer $d$, so the condition $a<p$ yields $$2^{\nu_2(\prod(p_i-1))}d\frac{\prod p_i-1}{\prod(p_i-1)}<\frac{p_j-1}{2^{\nu_2(p_j-1)}}$$ for some prime $p_j\mid\prod p_i$.
The second case $q=2$ implies that $n=2^kr=am+1$ where $m\mid\phi(r)$; that is, for some positive integer $g$ we have $g(2^kr-1)=a\phi(r)$.
The third case $q-1=v$ forces $m=\phi(r)$, so $m=1$. This is a contradiction as there is no prime $p$ that can divide $m$.
I had answered in a comment but it may be useful to know that for some values of $p$ and $n$ there are several exceptions, for example:
$p=13 \quad $ and $\quad n=19$
$19 \rightarrow 248 \rightarrow 31 \rightarrow 404\rightarrow 101\rightarrow 1314\rightarrow 73\rightarrow 950\rightarrow 19 $
Best Answer
I think your highlighted result is true. Here are some ideas for a proof:
First observe that the image of your function $f$ will be the same size (cardinality) as the image of the squaring function.
Second observe that since opposites in $\mathbb{Z}_n$ (i.e., $k$ and $n-k$) square to the same thing, and since only $0$ is its own opposite (since $n$ is odd); we have that $2|$ Im $f |\le n+1$, with equality only if at most two elements square to the same value; and only $0$ squares to $0$.
For $\Leftarrow)$: Suppose $n$ is an odd prime. it is well known that exactly half the nonzero elements in $\mathbb{Z}_n$ are quadratic residues (perfect squares) $\pmod{n}$. Also $0$ is a perfect square. The result follows.
For $\Rightarrow)$: Case 1: $n$ is a prime power. Say $n=p^k$ for some odd prime $p$ and some integer $k>1$. Then multiple elements will square to $0$ (certainly $0$ and $2\cdot p^{k-1}$ both square to $0$). So as noted above, (second observation), the result will hold.
Case 2: $n$ is divisible by (at least) two distinct odd primes. Then $x^2\equiv 1 \pmod{n}$ will have more than two solutions (this is a fairly standard result, that can be proved using the Chinese Remainder Theorem). So again the second observation above shows that the result will hold.