I don't think that analytic power series can be controlled by algebraic power series in an algebraic way, because they are often transcendental over the former.

Suppose an automorphism $\phi$ of $R$ is identity on the subring of algebraic power series. Is $\phi$ identity on the analytic power series?

The answer is no.

For a counter-example let $k$ be any field of characteristic zero and let $K \subseteq k((X))$ be the field of elements that are algebraic over $k(X)$. By the accepted answer of What's an example of a transcendental power series?, the exponental function $\text{exp}$ is transcendental over $k(X)$ and by transitivity of algebraic extensions, $\text{exp}$ is also transcendental over $K$. Extend the identity on $K$ to the automorphism $$\phi: K(\text{exp}) \to K(\text{exp}),\,\,\text{exp}\mapsto \text{exp} + X$$ (the inverse sends $\text{exp}\mapsto \text{exp} - X)$.

Let $R$ be an algebraic closure of $k((X))$. Then $\phi$ extends to an automorphism of $R$ (cf. https://en.wikipedia.org/wiki/Transcendence_degree, "Applications"). This $\phi$ is the identity on algebraic power series but not on analytic power series.

At least, does $\phi$ send analytic to analytic ?

I think the answer is no, but I have no example. To produce an example one could look for a non-analytic power series $f$ such that $\text{exp}, f$ are algebraically independent over $K$. Then modify the $\phi$ above by mapping $\text{exp} \mapsto f,\,\, f \mapsto \text{exp}$.

Added Apr 11, 2017: The answer to the second question is also no:

Let $k = \mathbb{C}$. As indicated above, it suffices to find a non-analytic power series $f$, that is algebraically independet over $K(\text{exp})$.

By the Newton-Puiseux Theorem (see http://www.emis.de/journals/UIAM/actamath/PDF/38-279-282.pdf), the field $P_a$ of analytic Puiseux series over $\mathbb{C}$ is algebraically closed. In particular, each non-analytic power series is transcendental over $P_a \supseteq K(\text{exp})$. So, we can take, for example, $f := \sum_{n\ge 1} n^n X^n$.

I don't think that analytic power series can be controlled by algebraic power series in an algebraic way, because they are often transcendental over the former.

Suppose an automorphism $\phi$ of $R$ is identity on the subring of algebraic power series. Is $\phi$ identity on the analytic power series?

The answer is no.

For a counter-example let $k$ be any field of characteristic zero and let $K \subseteq k((X))$ be the field of elements that are algebraic over $k(X)$. By the accepted answer of What's an example of a transcendental power series?, the exponental function $\text{exp}$ is transcendental over $k(X)$ and by transitivity of algebraic extensions, $\text{exp}$ is also transcendental over $K$. Extend the identity on $K$ to the automorphism $$\phi: K(\text{exp}) \to K(\text{exp}),\,\,\text{exp}\mapsto \text{exp} + X$$ (the inverse sends $\text{exp}\mapsto \text{exp} - X)$.

Let $R$ be an algebraic closure of $k((X))$. Then $\phi$ extends to an automorphism of $R$ (cf. https://en.wikipedia.org/wiki/Transcendence_degree, "Applications"). This $\phi$ is the identity on algebraic power series but not on analytic power series.

At least, does $\phi$ send analytic to analytic ?

I think the answer is no, but I have no example. To produce an example one could look for a non-analytic power series $f$ such that $\text{exp}, f$ are algebraically independent over $K$. Then modify the $\phi$ above by mapping $\text{exp} \mapsto f,\,\, f \mapsto \text{exp}$.

Added Apr 11, 2017: The answer to the second question is also no:

Let $k = \mathbb{C}$. As indicated above, it suffices to find a non-analytic power series $f$, that is algebraically independet over $K(\text{exp})$.

By the Newton-Puiseux Theorem (see http://www.emis.de/journals/UIAM/actamath/PDF/38-279-282.pdf), the field $P_a$ of analytic Puiseux series over $\mathbb{C}$ is algebraically closed. In particular, each non-analytic power series is transcendental over $P_a \supseteq K(\text{exp})$. So, we can take, for example, $f := \sum_{n\ge 1} n^n X^n$.

I don't think that analytic power series can be controlled by algebraic power series in an algebraic way, because they are often transcendental over the former.

Suppose an automorphism $\phi$ of $R$ is identity on the subring of algebraic power series. Is $\phi$ identity on the analytic power series?

The answer is no.

For a counter-example let $k$ be any field of characteristic zero and let $K \subseteq k((X))$ be the field of elements that are algebraic over $k(X)$. By the accepted answer of What's an example of a transcendental power series?, the exponental function $\text{exp}$ is transcendental over $k(X)$ and by transitivity of algebraic extensions, $\text{exp}$ is also transcendental over $K$. Extend the identity on $K$ to the automorphism $$\phi: K(\text{exp}) \to K(\text{exp}),\,\,\text{exp}\mapsto \text{exp} + X$$ (the inverse sends $\text{exp}\mapsto \text{exp} - X)$.

Let $R$ be an algebraic closure of $k((X))$. Then $\phi$ extends to an automorphism of $R$ (cf. https://en.wikipedia.org/wiki/Transcendence_degree, "Applications"). This $\phi$ is the identity on algebraic power series but not on analytic power series.

At least, does $\phi$ send analytic to analytic ?

I think the answer is no, but I have no example. To produce an example one could look for a non-analytic power series $f$ such that $\text{exp}, f$ are algebraically independent over $k(X)$. Then modify the $\phi$ above by mapping $\text{exp} \mapsto f,\,\, f \mapsto \text{exp}$.

I don't think that analytic power series can be controlled by algebraic power series in an algebraic way, because they are often transcendental over the former.

Suppose an automorphism $\phi$ of $R$ is identity on the subring of algebraic power series. Is $\phi$ identity on the analytic power series?

The answer is no.

For a counter-example let $k$ be any field of characteristic zero and let $K \subseteq k((X))$ be the field of elements that are algebraic over $k(X)$. By the accepted answer of What's an example of a transcendental power series?, the exponental function $\text{exp}$ is transcendental over $k(X)$ and by transitivity of algebraic extensions, $\text{exp}$ is also transcendental over $K$. Extend the identity on $K$ to the automorphism $$\phi: K(\text{exp}) \to K(\text{exp}),\,\,\text{exp}\mapsto \text{exp} + X$$ (the inverse sends $\text{exp}\mapsto \text{exp} - X)$.

Let $R$ be an algebraic closure of $k((X))$. Then $\phi$ extends to an automorphism of $R$ (cf. https://en.wikipedia.org/wiki/Transcendence_degree, "Applications"). This $\phi$ is the identity on algebraic power series but not on analytic power series.

At least, does $\phi$ send analytic to analytic ?

I think the answer is no, but I have no example. To produce an example one could look for a non-analytic power series $f$ such that $\text{exp}, f$ are algebraically independent over $K$. Then modify the $\phi$ above by mapping $\text{exp} \mapsto f,\,\, f \mapsto \text{exp}$.

Added Apr 11, 2017: The answer to the second question is also no:

Let $k = \mathbb{C}$. As indicated above, it suffices to find a non-analytic power series $f$, that is algebraically independet over $K(\text{exp})$.

By the Newton-Puiseux Theorem (see http://www.emis.de/journals/UIAM/actamath/PDF/38-279-282.pdf), the field $P_a$ of analytic Puiseux series over $\mathbb{C}$ is algebraically closed. In particular, each non-analytic power series is transcendental over $P_a \supseteq K(\text{exp})$. So, we can take, for example, $f := \sum_{n\ge 1} n^n X^n$.