All fields are domains; being a domain is not some extra property that a field may or may not have.
The Frobenius homomorphism is often called the Frobenius endomorphism, since "endomorphism" is a more specific term meaning "map from something to itself". However, I'll use "homomorphism" here to avoid confusion.
Let $K=\mathbb{F}_p(T)$, the rational functions in the variable $T$ over the field $\mathbb{F}_p=\mathbb{Z}/p\mathbb{Z}$. That is,
$$K=\mathbb{F}_p(T)=\left\{\,\frac{f}{g}\,\Bigg|\,\,\,f,g\in\mathbb{F}_p[T], g\neq0\right\}.$$
Then the Frobenius homomorphism $\phi:K\to K$ is not surjective, because (for instance) the element $T$ is not in the image of $\phi$. This is because if we had $\phi(\frac{f}{g})=\frac{f^{\;p}}{g^{\;p}}=T$ for some $\frac{f}{g}\in K$, then we have $p(\deg(f)-\deg(g))=\deg(T)=1$, which is impossible as $\deg(f)-\deg(g)$ is an integer and $1$ is not divisible by $p$.
Now let $K=\mathbb{F}_p$. Then the Frobenius homomorphism $\phi:K\to K$ is surjective, because it is in fact equal to the identity map, i.e. $\phi(x)=x$ for all $x\in K$. More generally, if $K$ is any finite field of characteristic $p$, then the Frobenius homomorphism is surjective, because (as I will show below) it is always injective, and an injective function from a finite set to itself must be surjective.
However, it is not necessary that $K$ be finite in order for the Frobenius homomorphism to be surjective. For example, now let $K=\mathbb{F}_p(T^{\;1/p^\infty})$. That is,
$$K=\mathbb{F}_p(T^{\;1/p^\infty})=\mathbb{F}_p(T,\sqrt[p]{T\;},\sqrt[p^2]{T\;},\ldots).$$ This is certainly an infinite field. The Frobenius homomorphism $\phi:K\to K$ is surjective. For example, the element $\alpha\in K$,
$$\alpha=\frac{\sqrt[p]{T\;}+2\cdot(\sqrt[p^2]{T\;})^3}{T^{\;2}-5T^{\;p}}$$
is in the image of $\phi$, because we can replace every occurrence of a $T^{\;p^d}$ with a one-lower-power-of-$p$ exponent, i.e.
$$\phi(\beta)=\alpha,\text{ where }\quad\beta=\frac{\sqrt[p^2]{T\;}+2\cdot(\sqrt[p^3]{T\;})^3}{(\sqrt[p]{T\;})^{\;2}-5T}$$
Similarly with any other element of $K$. (Note that $2$ or $5$ could very well be equal to $p$, and hence equal to $0$.)
Incidentally, the field $\mathbb{F}_p(T^{\;1/p^\infty})$ is called the perfection of the field $\mathbb{F}_p(T)$. It is a theorem that a field $K$ of characteristic $p$ is perfect if and only if the Frobenius homomorphism $\phi:K \to K$ is surjective, and by adding in $p^n$-th roots of every element of $\mathbb{F}_p(T)$, we have made a field for which it must be surjective, hence the name.
Now, some books (for example, McCarthy's Algebraic Extensions of Fields) use the term "isomorphism" to mean what we now call "monomorphism" (injective homomorphism), and when they want to express what we now call "isomorphism" (bijective homomorphism), they would say "onto isomorphism" (since onto is a synonym for surjective, and bijective = injective + surjective).
If for some reason you are using this older terminology, then the Frobenius homomorphism for a field is always an "isomorphism" (i.e. injective). This is because, if $K$ is any field of characteristic $p$, and $\phi:K\to K$ is the Frobenius homomorphism, then
$$\begin{align*}\phi(\alpha)=\alpha^p=\beta^p=\phi(\beta)&\implies\alpha^p-\beta^p=0\\ &\implies(\alpha-\beta)^p=0\\ & \implies\alpha-\beta=0\\ & \implies\alpha=\beta.\end{align*}$$
In fact, any homomorphism from a field $K$ to a non-zero ring $R$ must be injective (what could its kernel be, as an ideal of the field $K$?) Also note that, by the First Isomorphism Theorem for Rings, we therefore have that any homomorphism $f:K\to R$ from a field $K$ to a non-zero ring $R$ is an isomorphism (in the modern sense) onto the subring of $R$ that is the image of $f$, i.e. if we threw out the rest of $R$ and only looked at the subring $f(K)$, then $f:K\to f(K)$ is an isomorphism (in the modern sense). This is what Jyrki Lahtonen conjectured that you meant above.
If we look at things that are not domains (and so in particular are not fields), then the Frobenius homomorphism need not be injective. For example, let $R=\mathbb{F}_p[x]/(x^p)$. That is,
$$R=\mathbb{F}_p[x]/(x^p)=\{a_0+\cdots+a_{p-1}x^{p-1}+(x^p)\mid a_i\in\mathbb{F}_p\}.$$
Then if $\phi:R\to R$ is the Frobenius homomorphism, we have for example $x+(x^p)\neq0+(x^p)$, but $$\phi(x+(x^p))=x^p+(x^p)=0+(x^p)=0^p+(x^p)=\phi(0+(x^p)).$$
On a final note, I just want to emphasize that the concept of a Frobenius homomorphism is only applicable if the field is of characteristic $p$ (as all the examples above were). For example, $\mathbb{Q}$ is a field of characteristic $0$, and for any prime $p$, the function $f:\mathbb{Q}\to\mathbb{Q}$ defined by $f(a)=a^p$ is not a homomorphism (compare $f(2)$ with $2\cdot f(1)$).
Fill in the (usually small) details in the following:
It is probably easier to show the double negation: $\;\Bbb F\;$ is inseparable iff $\;\Bbb F^p\setminus\Bbb F\neq\emptyset\;$
== So take $\;a\in\Bbb F^p\setminus\Bbb F\;$ . If $\;\beta\;$ in some non-trivial extension of $\;\Bbb F\;$ is s.t. $\;\beta^p=a\;$ , then
$$f_a(x):=x^p-a=x^p-\beta^p=(x-\beta)^p\in\Bbb F[x]$$
If $\;f_a(x)\;$ isn't irreducible over $\;\Bbb F\;$, then the right hand above shows that it must have a non-trivial factor in $\;\Bbb F[x]\;$ of the form $\;(x-\beta)^n\;,\;\;1\le n\le p-1\;$ . This polynomial's coefficient of $\;x^{n-1}\;$ is $\;-n\beta\;$ and it belongs to $\;\Bbb F\;$ . Since $\; n(\neq0)\;$ belongs to the prime field $\;\Bbb F_p\le\Bbb F\;$ , we get that it is obviously invertible there and thus
$$-n\beta\in\Bbb F\implies \beta\in\Bbb F\;\implies\;a=\beta^p\in\Bbb F$$
which is a contradiction. Thus, $\;f_a(x)\;$ is irreducible.
== Suppose now there exists an inseparable irreducible $\;g(x)\in\Bbb F[x]\;$ . Then it must be that $\;g'(x)\equiv0\;$ , so that $\;g\;$ is a polynomial in $\;x^p\;$ :
$$g(x)=\sum_{k=0}^m a_kx^{pk}$$
If we had that $\;\Bbb F^p=\Bbb F\;$ , then for all $\;0\le k\le m\;$ there exist $\;b_k\;$ s.t. $\;a_k=b_k^p\;$ , but then
$$g(x)=\sum_{k=0}^m a_kx^{pk}=\sum_{k=0}^mb_k^p\left(x^k\right)^p=\left(\sum_{k=0}^m b_kx^k\right)^p$$
which contradicts the fact that $\;g\;$ is irreducible.
Best Answer
Let $f(x)$ be an irreducible polynomial. We know that $\gcd(f,f')\neq 1$ only if $f'=0$, which would mean that $f(x)$ is a polynomial in $x^p$; that is, $$f(x) = a_nx^{p^n} + a_{n-1}x^{p^{n-1}}+\cdots+a_1x^p + a_0.$$ So it suffices to show that no polynomial in $x^p$ is irreducible over $K$.
So let $f(x) = a_nx^{p^n}+\cdots + a_1x^p + a_0$ be a polynomial in $x^p$ with coefficients in $K$.
Since the Frobenius endomorphism is surjective, for each $k=0,\ldots,n$ there exists $b_k\in K$ such that $b_k^p=a_k$. So we can write $$\begin{align*} f(x) &= a_nx^{p^n}+\cdots + a_1x^p + a_0\\ &= b_n^px^{p^n}+\cdots + b_1^px^p + b_0^p\\ &= (b_nx^{p^{n-1}})^p+(b_{n-1}x^{p^{n-2}})^p+\cdots + (b_1x)^p + b_0^p\\ &= (b_nx^{p^{n-1}} + \cdots + b_1x + b_0)^p. \end{align*}$$ Thus, $f(x)$ is not irreducible, as desired.
Therefore, if $f$ is irreducible, then $\gcd(f,f')=1$, which means $f$ does not have multiple roots; i.e., $f$ is separable over $K$.